Explanation:
While working in a laboratory it is necessary to be free from any kind of distraction. This is because a distraction while working in a laboratory can lead to serious consequences.
For this it is important to put all your electronic devices like mobile phones, tablets etc in your backpack. Also, you should not carry any kind of eatable, water bottle or anything into the lab.
This is because chemicals do react with food items. Hence, then food does not remain fit for consumption.
Thus, we can conclude that pen or pencil and lab manual personal items are permitted to be at your workstation in the organic laboratory.
An aluminum atom has a mass of 4.48 * 10-23 g and a small airplane has a mass of 5000 kg. Use this information to answer the questions below. Be sure your answers have the correct number of significant digits.(a) What is the mass of 1 mole of aluminum atoms?(b) How many moles of aluminum atoms have a mass equal to the mass of a small airplane?
Answer:
a. 27g/mol
b. 1.85 x 10^5 moles
Explanation:Please see attachment for explanation
A chemist added an excess of sodium sulfate to a solution of a soluble barium compound to precipitate all of the barium ions as barium sulfate, BaSO4. How many grams of barium ions are in a 441-mg sample of the barium compound if a solution of the sample gave 403 mg BaSO4 precipitate? What is the mass percentage of barium compound?
Answer:
259.497 mg, 58.84%
Explanation:
BaSO₄ → Ba²⁺ + SO₄²⁻
to calculate the mole of BaSO₄
mole BaSO₄ = mass given / molar mass = 403 mg / 233.38 g/mol = 1.7268 mol
comparing the mole ratio
1.7268 mol of BaSO₄ yields 1.7268 mol of Ba²⁺
403 mg BaSO₄ yields ( 1.7268 × 137.327 ) where 137.327 is the molar mass of Barium mol of Ba²⁺
441 mg BaSO₄ will yield ( 1.7268 × 137.327 × 441 mg ) / 403 mg = 259 .497 mg
mas percentage of the Barium compound = 259 .497 mg / 441 mg × 100 = 58.84%
Whay is it important for a forensic scientist to have as much information as possible about the place where a possible drug substance was found?
Answer: it is important because the location of crime tells a lot about possible substances that may have been in the area, may affect the drug substance in question, thus helping in tracking location other substances, why they are there, who used or placed them, assisting you in resolving a scene of crime
Explanation:
a. Determine the sign of the charge on each rod.
b.Compute the ratio of the linear charge densities on the two rods, .
Answer.
The electric field due to a negative charge points radially in from all directions (because a positive test charge placed near it would feel a force pointing toward it). Generally, electric field lines always point from positive charges and toward negative charges.
Your question I s incomplete, please endeavor you update your question with the necessary parameters.
determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.
This is an incomplete question, here is a complete question.
Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.
(a) [tex]1s^22s^22p^63s^23p^5[/tex]
(b) [tex]1s^22s^22p^63s^23p^63d^74s^2[/tex]
(c) [tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^6[/tex]
(d) [tex]1s^22s^22p^63s^23p^63d^4s^1[/tex]
Answer :
(a) [tex]1s^22s^22p^63s^23p^5[/tex] → Halogen
(b) [tex]1s^22s^22p^63s^23p^63d^74s^2[/tex] → Transition metal
(c) [tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^6[/tex] → Transition metal
(d) [tex]1s^22s^22p^63s^23p^63d^4s^1[/tex] → Transition metal
Explanation :
Inert gas : These are the gases which lie in group 18.
Their general electronic configuration is: [tex]ns^2np^6[/tex] where n is the outermost shell.
Halogen : These are the elements which lie in group 17.
Their general electronic configuration is: [tex]ns^2np^5[/tex] where n is the outermost shell.
An alkali metal : These are the elements which lie in group 1.
Their general electronic configuration is: [tex]ns^1[/tex] where n is the outermost shell.
An alkaline earth metal : These are the elements which lie in group 2.
Their general electronic configuration is: [tex]ns^2[/tex] where n is the outermost shell.
Transition elements : They are the elements which lie between 's' and 'p' block elements. These are the elements which lie in group 3 to 12. The valence electrons of these elements enter d-orbital.
Their general electronic configuration is: [tex](n-1)d^{1-10}ns^{0-2}[/tex] where n is the outermost shell.
(a) [tex]1s^22s^22p^63s^23p^5[/tex]
The element having this electronic configuration belongs to the halogen family.
(b) [tex]1s^22s^22p^63s^23p^63d^74s^2[/tex]
The element having this electronic configuration belongs to the transition family.
(c) [tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^6[/tex]
The element having this electronic configuration belongs to the transition family.
(d) [tex]1s^22s^22p^63s^23p^63d^4s^1[/tex]
The element having this electronic configuration belongs to the transition family.
1. [tex][Ar]4s^2 3d^10 4p^6[/tex] is an inert gas (krypton, Kr).
2. [tex][Kr]5s^2 4d^10 5p^2[/tex] is a transition metal (zinc, Zn).
3. [tex][He]2s^1[/tex] is an alkali metal (lithium, Li).
4. [tex][Ne]3s^2 3p^5[/tex] is a halogen (chlorine, Cl).
5. [tex][Ar]4s^2[/tex] is an alkaline earth metal (calcium, Ca).
6. [tex][Xe]6s^2 4f^14 5d^10 6p^3[/tex] is not a transition metal but a post-transition metal (bismuth, Bi).
To determine the type of element based on its electron configuration, one must understand the characteristic configurations of the different groups of elements in the periodic table. Here are some general rules:
- Inert gases (also known as noble gases) have completely filled s and p subshells in their valence shell. Their configurations typically end in [tex]ns^2 np^6[/tex], except for helium, which has [tex]1s^2[/tex].
- Halogens have the [tex]ns^2 np^5[/tex] configuration in their valence shell, with the exception of helium, which has [tex]1s^2[/tex].
- Alkali metals have a single electron in their outermost s subshell, denoted as [tex]ns^1[/tex].
- Alkaline earth metals have two electrons in their outermost s subshell, denoted as [tex]ns^2[/tex].
- Transition metals have incompletely filled d subshells in their penultimate energy level. Their configurations typically start with [tex](n-1)d^1 to (n-1)d^10[/tex] in the valence shell.
Let's analyze the given electron configurations:
1. [tex][Ar]4s^2 3d^10 4p^6[/tex]
- This configuration has a filled 4s subshell, a filled 3d subshell (10 electrons), and a filled 4p subshell. It corresponds to the configuration of an inert gas, specifically krypton (Kr), which has the full configuration [tex][Ar]4s^2 3d^10 4p^6[/tex].
2. [tex][Kr]5s^2 4d^10 5p^2[/tex]
- This configuration has a filled 5s subshell and a filled 4d subshell, but the 5p subshell is not filled (only 2 electrons out of a possible 6). This does not correspond to an inert gas, a halogen, an alkali metal, or an alkaline earth metal. It is a configuration of an element in group 12, which is a transition metal. Specifically, this is the configuration for zinc (Zn).
3. [tex][He]2s^1[/tex]
- This configuration has a single electron in the 2s subshell, which is characteristic of an alkali metal. Specifically, this is the configuration for lithium (Li).
4. [tex][Ne]3s^2 3p^5[/tex]
- This configuration has a filled 3s subshell and the 3p subshell is missing one electron to be filled (5 electrons out of a possible 6). This is the configuration of a halogen, specifically chlorine (Cl).
5. [tex][Ar]4s^2[/tex]
- This configuration has a filled 4s subshell, which is characteristic of an alkaline earth metal. Specifically, this is the configuration for calcium (Ca).
6. [tex][Xe]6s^2 4f^14 5d^10 6p^3[/tex]
- This configuration has a filled 6s subshell, a filled 4f subshell (14 electrons), and a filled 5d subshell (10 electrons), but the 6p subshell is not filled (only 3 electrons out of a possible 6). This does not correspond to an inert gas, a halogen, an alkali metal, or an alkaline earth metal. It is the configuration of an element in group 13, which is not a transition metal but a post-transition metal. Specifically, this is the configuration for bismuth (Bi).
In summary:
1. [tex][Ar]4s^2 3d^10 4p^6[/tex] is an inert gas (krypton, Kr).
2. [tex][Kr]5s^2 4d^10 5p^2[/tex] is a transition metal (zinc, Zn).
3. [tex][He]2s^1[/tex] is an alkali metal (lithium, Li).
4. [tex][Ne]3s^2 3p^5[/tex] is a halogen (chlorine, Cl).
5. [tex][Ar]4s^2[/tex] is an alkaline earth metal (calcium, Ca).
6.[tex][Xe]6s^2 4f^14 5d^10 6p^3[/tex] is not a transition metal but a post-transition metal (bismuth, Bi).
hcl A solution of sodium thiosulfate (Na2S2O3) in water is 12.00 % sodium thiosulfate by mass. Its density is 1.1003 g cm-3 at 20 °C. Compute its molarity, mole fraction, and molality at this temperature.
Answer:
[Na₂S₂O₃] = 0.83 M
[Na₂S₂O₃] = 0.86 m
Mole fraction = 0.015
Explanation:
Na₂S₂O₃ 12 % by mass. This data means, that 12 g of solute are contained in 100 g of solution.
Let's find out the volume of solution, with density to determine molarity.
Solution density = Solution mass / Solution volume
1.1003 g/cm³ = 100 g / Solution volume
100 g / 1.1003 g/cm³ = Solution volume → 90.88 mL (1cm³ = 1mL)
Now, that we have volume, we can calculate molarity
Molarity is mol/L
90.88 mL = 0.09088 L
12 g / 158.12 g/mol = 0.0759 moles
0.0759 moles / 0.09088 L = 0.83 M
Total mass of solution = 100 g
12 g + Solvent mass = 100 g
Solvent mass = 100 g - 12 g → 88 g
Molality = moles of solute /1kgof solvent
88 g = 0.088 kg
0.0759 moles / 0.088 kg = 0.86 m
As solvent mass is 88 g, let's determine solvent's moles for mole fraction
88 g / 18 g/mol = 4.89 moles
Mole fraction = moles of solute / moles of solutes + moles of solvent
Mole fraction = 0.0759 mol / 0.0759 mol + 4.89 moles = 0.015
Phosphorus is obtained primarily from ores containing calcium phosphate.
If a particular ore contains 53.1% calcium phosphate, what minimum mass of the ore must be processed to obtain 3.57 kg of phosphorus?
Answer:
33.7 kg
Explanation:
Let's consider calcium phosphate Ca₃(PO₄)₂.
The molar mass of Ca₃(PO₄)₂ is 310.18 g/mol and the molar mass of P is 30.97 g/mol. In 1 mole of Ca₃(PO₄)₂ (310.18 g) there are 2 × 30.97 g = 61.94 g of P. The mass of Ca₃(PO₄)₂ that contains 3.57 kg (3.57 × 10³ g) of P is:
3.57 × 10³ g × (310.18 g Ca₃(PO₄)₂/61.94 g P) = 1.79 × 10⁴ g Ca₃(PO₄)₂
A particular ore contains 53.1% calcium phosphate. The mass of the ore that contains 1.79 × 10⁴ g of Ca₃(PO₄)₂ is:
1.79 × 10⁴ g Ca₃(PO₄)₂ × (100 g Ore/ 53.1 g Ca₃(PO₄)₂) = 3.37 × 10⁴ g Ore = 33.7 kg Ore
Each of the following substrates can react with a nucleophile in a substitution reaction. Select the substrate that cannot undergo substitution via neighboring group participation (NGP). A B C D
Answer:
Substrate D
Explanation:
In substitution reactions the tertiary substrates cannot undergo substitution via neighboring group participation (NGP) due to the steric impediment, this means that the volume occupied by the substituents is very large and makes it impossible to attack the nucleophile to the substrate carbon.
Chlorine is the most widely used disinfectant for killing pathogens during water treatment. Determine the kilograms of chlorine used daily at a water treatment plant handling 10,000 cubic meters per day of flow at a chlorine dosage of 10 mg/L. [Hint: Multiply the flow rate by the chlorine dosage and make appropriate conversions.] Ans
Explanation:
The given data is as follows.
Flow of chlorine (Q) = [tex]10,000 m^{3}/day[/tex]
Amount of liter present per day is as follows.
[tex]10,000 \times 10^{3} l/day[/tex]
It is given that dosage of chlorine will be as follows.
10 mg/l = [tex]10 \times 10^{-6}[/tex] kg/l
Therefore, total chlorine requirement is as follows.
Total chlorine requirement = [tex](10,000 \times 10^{3}) \times (10 \times 10^{-6})[/tex] kg/day
= 100 kg/day
Thus, we can conclude that the kilograms of chlorine used daily at the given water treatment plant is 100 kg/day.
A gasoline spill is approximately 4 liters of liquid. What volume of vapor is created at 1 bar and 293 K when the liquid evaporates? The density of regular gasoline can be estimated by treating it as pure isooctane (2,2,4-trimethylpentane rhoL = 0.692 g/cm3) at 298 K and 1 bar.
Answer:
[tex]V=591.748 L[/tex]
Explanation:
Assumption:
Ideal Vapors/Ideal gas
Formula for ideal Gas:
[tex]PV=nR_uT[/tex]
Where:
P is the pressure
V is the Volume
n is the number of moles = m/M
R_u is Universal Gas Constant=0.08314 L*bar/(K*mol)
T is the temperature in Kelvin
Calculating Number of moles n:
n=Mass/Molar Mass
[tex]Mass=\rho_L*Volume\\Mass=0.692*(4000 cm^3)........... (4 liter * 1000cm^3/Liters =4000 cm^3)\\Mass=2768 g[/tex]
Molar Mass of gasoline=114g/mol
[tex]n=\frac{2768}{114} \\n=24.2807 moles[/tex]
Now:
[tex]PV=nR_uT[/tex]
[tex]V=\frac{nR_uT}{P}\\V=\frac{24.2807*0.08314*293}{1 bar}\\V=591.748 L[/tex]
How many electrons in an atom can have each of the following quantum number or sublevel designations? (a) 4p (b) n = 3, l = 1, m1 = +1 (c) n = 5, l = 3
Answer:
a) Six electrons
b) Two electrons
c) Fourteen electrons
Explanation:
n is the principal quantum number and defines the energy level of orbital. The shape of the orbital is described by azimuthal quantum number (l) and it also determine the angular momentum. It values give the following information
l = 0, define s orbital (single orbital)
l = 1, define p orbitals (three orbitals)
l = 2, define d orbitals (five orbitals)
l = 3, define f orbitals (seven orbitals)
These are further specified by magnetic quantum number (ml) which gives the orientation of the orbital. Its value ranges from +1 to -1, for example ml value of five d orbitals are +2, +1, 0, -1, -2. From this information we can predict the number of electrons that will have the given sub-level designations
a) n = 4 and orbital is p, there are three p orbitals as the ml is not defined, so six electrons will have this quantum number
b) In this part, the orbital is defined i.e. ml = +1. A single orbital can have only two electrons, so these electrons will have the given quantum number.
c) l = 3, is for f orbital, which have seven orbitals. The total number of electrons in it is fourteen. All of these electrons will have this quantum number.
(a) An atom with the quantum number of 4p will have 6 electrons.
(b) An atom with the quantum number of n=3, i = 1, m1 = +1 will have 2 electrons.
(c) An atom with the quantum number of n=5, i = 3, will have 14 electrons.
The number of electrons an atom in the given quantum number can have is calculated as follows;
(a) 4p --- p-orbital has 3 sub-shells and the atom will have maximum of 6 electrons.
(b) n = 3, l = 1, m1 = +1 --- this corresponds to 3p - orbital and each atom will have maximum of 2 electrons.
(c) l = 3, corresponds to f - orbital
f-orbitals have 7 sub-shell and the atom will have maximum of 14 electrons.
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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).
The question is incomplete, here is the complete question:
Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).
Atmospheric Gas Mole Fraction kH mol/(L*atm)
[tex]N_2[/tex] [tex]7.81\times 10^{-1}[/tex] [tex]6.70\times 10^{-4}[/tex]
[tex]O_2[/tex] [tex]2.10\times 10^{-1}[/tex] [tex]1.30\times 10^{-3}[/tex]
Ar [tex]9.34\times 10^{-3}[/tex] [tex]1.40\times 10^{-3}[/tex]
[tex]CO_2[/tex] [tex]3.33\times 10^{-4}[/tex] [tex]3.50\times 10^{-2}[/tex]
[tex]CH_4[/tex] [tex]2.00\times 10^{-6}[/tex] [tex]1.40\times 10^{-3}[/tex]
[tex]H_2[/tex] [tex]5.00\times 10^{-7}[/tex] [tex]7.80\times 10^{-4}[/tex]
Answer: The solubility of hydrogen gas in water at given atmospheric pressure is [tex]1.48\times 10^{-10}M[/tex]
Explanation:
To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:
[tex]p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}[/tex]
where,
[tex]p_A[/tex] = partial pressure of hydrogen gas = ?
[tex]p_T[/tex] = total pressure = 0.380 atm
[tex]\chi_A[/tex] = mole fraction of hydrogen gas = [tex]5.00\times 10^{-7}[/tex]
Putting values in above equation, we get:
[tex]p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm[/tex]
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{H_2}=K_H\times p_{H_2}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]7.80\times 10^{-4}mol/L.atm[/tex]
[tex]p_{H_2}[/tex] = partial pressure of hydrogen gas = [tex]1.9\times 10^{-7}atm[/tex]
Putting values in above equation, we get:
[tex]C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M[/tex]
Hence, the solubility of hydrogen gas in water at given atmospheric pressure is [tex]1.48\times 10^{-10}M[/tex]
Consider this reaction:
2Cl2O5 (g) → 2Cl2 (g) + 5O2 (g)
At a certain temperature it obeys this rate law.
rate = (6.48 M-1 • s-1)[Cl2O5]2
Suppose a vessel contains Cl2O5 at a concentration of 1.16 M. Calculate the concentration of Cl2O5 in the vessel 0.820 seconds later. You may assume no other reaction is important.
Round your answer to 2 significant digits.
Answer : The concentration of [tex]Cl_2O_5[/tex] in the vessel 0.820 seconds later is, 0.16 M
Explanation :
The given reaction is:
[tex]2Cl_2O_5(g)\rightarrow 2Cl_2(g)+5O_2(g)[/tex]
The rate law expression is:
[tex]rate=(6.48M^{-1}s^{-1})[Cl_2O_5]^2[/tex]
The expression used for second order kinetics is:
[tex]kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}[/tex]
where,
k = rate constant = [tex]6.48M^{-1}s^{-1}[/tex]
t = time = 0.820 s
[tex][A_t][/tex] = final concentration = ?
[tex][A_o][/tex] = initial concentration = 1.16 M
Now put all the given values in the above expression, we get:
[tex]6.48\times 0.820=\frac{1}{[A_t]}-\frac{1}{1.16}[/tex]
[tex][A_t]=0.16M[/tex]
Therefore, the concentration of [tex]Cl_2O_5[/tex] in the vessel 0.820 seconds later is, 0.16 M
A student needs to convert a volume of 3 Tbsp (tablespoons) to mL. The student finds out that 1 tablespoon is equivalent to 14.79 mL. Choose the correct numbers and units that must be found in the conversion factor to solve this problem.
Answer:
The conversion factor is 14.79 mL/Tbsp.
Explanation:
To do an unity conversiton, we can make a factor by a ratio transformation:
[tex]3 Tbsp * \frac{14.79 mL}{1Tbsp}[/tex]
So, the conversion factor is 14.79 mL/Tbsp and 3 Tbsp has 44.37 mL.
The conversion factor of the tablespoon to volume has been 14.79mL/tablespoon.
Conversion factor has been defined as the unit that has been used to convert one unit range to another.
The computation with the conversion factor has been used in the system for the calculations and transform in the reactions.
Conversion factor for tablespoon to mLThe estimated value of 1 tablespoon has been found to be 14.79 mL. The conversion of 3 tablespoons to mL has been given as:
[tex]\rm 1\;tablespoons=14.79 \;mL\\3\;table spoons=3\;\times\;14.79 \;mL\\3\;tablespoons=44.37\;mL[/tex]
The volume of 3 tablespoon solution has been 44.37 mL.
The conversion has been performed with 1 tablespoon equivalent to 14.79 mL.
Thus, the conversion factor of the tablespoon to volume has been 14.79mL/tablespoon.
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Suppose 2.19g of barium acetate is dissolved of 15oml barium of a 0.10M acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it. Be sure your answer has the correct number of significant digits.
The question is incomplete, here is the complete question:
Suppose 2.19 g of barium acetate is dissolved in 150 mL of a 0.10M of aqueous solution of sodium chromate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it. Be sure your answer has the correct number of significant digits.
Answer: The final molarity of acetate ion in the solution is 0.12 M
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}[/tex] .....(1)
For Sodium chromate:Molarity of sodium chromate solution = 0.10 M
Volume of solution = 150 mL
Putting values in equation 1, we get:
[tex]0.10M=\frac{\text{Moles of sodium chromate}\times 1000}{150}\\\\\text{Moles of sodium chromate}=\frac{(0.10\times 150)}{1000}=0.015mol[/tex]
For barium acetate:To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of barium acetate = 2.19 g
Molar mass of barium acetate = 255.43 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of barium acetate}=\frac{2.19g}{255.43g/mol}=0.0086mol[/tex]
The chemical equation for the reaction of barium acetate and sodium chromate follows:
[tex]Ba(CH_3CO_2)_2+Na_2CrO_4\rightarrow BaCrO_4(s)+2Na^+(aq.)+2CH_3CO_2^-(aq.)[/tex]
By stoichiometry of the reaction:
1 mole of barium acetate reacts with 1 mole of sodium chromate
So, 0.0086 moles of barium acetate will react with = [tex]\frac{1}{1}\times 0.0086mol[/tex] of sodium chromate
As, given amount of sodium chromate is more than the required amount. So, it is considered as an excess reagent.
Thus, barium acetate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of barium acetate produces 2 moles of acetate ions
So, 0.0086 moles of barium acetate will produce = [tex]\frac{2}{1}\times 0.0086=0.0172mol[/tex] of acetate ion
Now, calculating the molarity of acetate ions in the solution by using equation 1:
Moles of acetate ion = 0.0172 moles
Volume of solution = 150 mL
Putting values in equation 1, we get:
[tex]\text{Molarity of acetate ions}=\frac{0.0172\times 1000}{150}\\\\\text{Molarity of acetate ions}=0.12M[/tex]
Hence, the final molarity of acetate ion in the solution is 0.12 M
Which of the following aqueous solutions contains the greatest number of ions?
a. 400. mL of 0.10 M NaCl;
b. 300. mL of 0.10 M CaCl2;
c. 200. mL of 0.10 M FeCl3
Answer : The correct option is, (b) 300. mL of 0.10 M CaCl₂
Explanation :
We have to calculate the number of moles of ions for the following aqueous solutions.
(a) 400. mL of 0.10 M NaCl
[tex]NaCl\rightarrow Na^++Cl^-[/tex]
NaCl dissociates to give 2 moles of ions.
[tex]\text{Moles of ion}=\text{Molarity of }NaCl\times \text{Volume of solution (in L)}[/tex]
Volume of solution = 400. mL = 0.400 L
[tex]\text{Moles of ion}=0.10M\times 0.400L=0.04mol[/tex]
Total moles of ion = 0.04 mol × 2 = 0.08 mol
(b) 300. mL of 0.10 M CaCl₂
[tex]CaCl_2\rightarrow Ca^{2+}+2Cl^-[/tex]
CaCl₂ dissociates to give 3 moles of ions.
[tex]\text{Moles of ion}=\text{Molarity of }CaCl_2\times \text{Volume of solution (in L)}[/tex]
Volume of solution = 300. mL = 0.300 L
[tex]\text{Moles of ion}=0.10M\times 0.300L=0.03mol[/tex]
Total moles of ion = 0.03 mol × 3 = 0.09 mol
(c) 200. mL of 0.10 M FeCl₃
[tex]FeCl_3\rightarrow Fe^{3+}+3Cl^-[/tex]
FeCl₃ dissociates to give 4 moles of ions.
[tex]\text{Moles of ion}=\text{Molarity of }FeCl_3\times \text{Volume of solution (in L)}[/tex]
Volume of solution = 200. mL = 0.200 L
[tex]\text{Moles of ion}=0.10M\times 0.200L=0.02mol[/tex]
Total moles of ion = 0.02 mol × 4 = 0.08 mol
From this we conclude that, [tex]CaCl_2[/tex] aqueous solutions contains the greatest number of ions.
Hence, the correct option is, (b) 300. mL of 0.10 M CaCl₂
The aqueous solution that contains the greatest number of ions is b. 300. mL of 0.10 M CaCl₂.
a. The number of moles in 400. mL of 0.10 M NaCl is:
[tex]0.400 L \times \frac{0.10mol}{L} = 0.040mol[/tex]
Each mole of NaCl contains 2 moles of ions (1 Na⁺ and 1 Cl⁻). The moles of ions in 0.040 moles of NaCl are:
[tex]0.040molNaCl \times \frac{2molIons}{1molNaCl} = 0.080 mol Ions[/tex]
b. The number of moles in 300. mL of 0.10 M CaCl₂ is:
[tex]0.300 L \times \frac{0.10mol}{L} = 0.030mol[/tex]
Each mole of CaCl₂ contains 3 moles of ions (1 Ca²⁺ and 2 Cl⁻). The moles of ions in 0.030 moles of CaCl₂ are:
[tex]0.030molNaCl \times \frac{3molIons}{1molNaCl} = 0.090 mol Ions[/tex]
c. The number of moles in 200. mL of 0.10 M FeCl₃ is:
[tex]0.200 L \times \frac{0.10mol}{L} = 0.020mol[/tex]
Each mole of FeCl₃ contains 4 moles of ions (1 Fe³⁺ and 3 Cl⁻). The moles of ions in 0.020 moles of FeCl₃ are:
[tex]0.020molNaCl \times \frac{4molIons}{1molNaCl} = 0.080 mol Ions[/tex]
The aqueous solution that contains the greatest number of ions is b. 300. mL of 0.10 M CaCl₂.
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A solution is made by mixing 36. g of heptane (C_7H_16) and 16. g of chloroform (CHCl_3). Calculate the mole fraction of heptane in this sol'. Be sure your answer has the correct number of significant digits.
Answer:
Mole fraction of solute (heptane) → 0.73
Explanation:
Mole fraction = Moles of solute or solvent / Total moles
Let's calculate the moles of everything:
Moles of solute → Mass of solute / Molar mass
36 g / 100 g/mol = 0.36 moles
Moles of solvent → Mass of solvent / Molar mass
16 g / 119.35 g/mol = 0.134 moles
Total moles = 0.36 + 0.134 = 0.494 moles
Mole fraction of solute = 0.36 / 0.494 → 0.73
The empirical formula of a compound is CH. At 200 degree C, 0.145 g of this compound occupies 97.2 mL at a pressure of 0.74 atm. What is the molecular formula of the compound
Answer:
The molecular formula = [tex]C_{6}H_{6}[/tex]
Explanation:
Given that:
Mass of compound, m = 0.145 g
Temperature = 200 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (200 + 273.15) K = 473.15 K
V = 97.2 mL = 0.0972 L
Pressure = 0.74 atm
Considering,
[tex]n=\frac{m}{M}[/tex]
Using ideal gas equation as:
[tex]PV=\frac{m}{M}RT[/tex]
where,
P is the pressure
V is the volume
m is the mass of the gas
M is the molar mass of the gas
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the values in the above equation as:-
[tex]0.74\times 0.0972=\frac{0.145}{M}\times 0.0821\times 473.15[/tex]
[tex]M=78.31\ g/mol[/tex]
The empirical formula is = [tex]CH[/tex]
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus,
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 12 + 1 = 13 g/mol
Molar mass = 78.31 g/mol
So,
Molecular mass = n × Empirical mass
78.31 = n × 13
⇒ n ≅ 6
The molecular formula = [tex]C_{6}H_{6}[/tex]
The molecular formula of the compound CH is [tex]C_6H_6[/tex].
Given:
P = 0.74 atm
V = 97.2 mL = 0.0972 L (since 1 mL = 0.001 L)
T = 200°C + 273.15 = 473.15 K (to convert from Celsius to Kelvin)
R = 0.0821 L·atm/(mol·K)
First, we solve for n, the number of moles:
[tex]\[ n = \frac{PV}{RT} \][/tex]
[tex]\[ n = \frac{(0.74 \text{ atm})(0.0972 \text{ L})}{(0.0821 \text{ L·atm/(mol·K)})(473.15 \text{ K})} \][/tex]
[tex]\[ n \approx \frac{(0.74)(0.0972)}{(0.0821)(473.15)} \][/tex]
[tex]\[ n \approx \frac{0.072048}{38.84015} \][/tex]
[tex]\[ n \approx 0.001855 \text{ mol} \][/tex]
Next, we calculate the molar mass (M) using the given mass (m) of the compound: [tex]\[ M = \frac{m}{n} \][/tex]
[tex]\[ M = \frac{0.145 \text{ g}}{0.001855 \text{ mol}} \][/tex]
[tex]\[ M \approx \frac{0.145}{0.001855} \][/tex]
[tex]\[ M \approx 78.17 \text{ g/mol} \][/tex]
The empirical formula mass of CH is:
[tex]\[ (1 \times 12.01 \text{ g/mol}) + (1 \times 1.008 \text{ g/mol}) = 13.018 \text{ g/mol} \][/tex]
To find the molecular formula, we divide the molar mass by the empirical formula mass:
[tex]\[ \text{Molecular formula mass} = n \times \text{Empirical formula mass} \][/tex]
[tex]\[ \text{Molecular formula mass} = n \times \text{Empirical formula mass} \][/tex]
[tex]\[ n = \frac{78.17}{13.018} \][/tex]
[tex]\[ n \approx 6 \][/tex]
Therefore, the molecular formula is 6 times the empirical formula, which is (CH)†. However, since the empirical formula, CH already represents one carbon and one hydrogen atom, the molecular formula is simply C†H†.
Upon reviewing the problem, it appears there was an error in the calculation of n, the number of moles. Let's correct this:
[tex]\[ n = \frac{(0.74)(0.0972)}{(0.0821)(473.15)} \][/tex]
[tex]\[ n = \frac{(0.74)(0.0972)}{(0.0821)(473.15)} \][/tex]
[tex]\[ n = \frac{(0.74)(0.0972)}{(0.0821)(473.15)} \][/tex]
This value of n is correct, and the subsequent calculations are also correct. However, the final molecular formula should be CH, not C†H†, because the correct calculation for n (the multiplier) is:
[tex]\[ n = \frac{78.17}{13.018} \][/tex]
[tex]\[ n = \frac{78.17}{13.018} \][/tex]
Since the empirical formula, CH is CH‚ multiplying by 6 gives us C†H†. However, we must consider that the subscripts in the empirical formula are the smallest whole-number ratio of atoms in the compound. Therefore, the correct molecular formula is obtained by multiplying the subscripts in the empirical formula by the same number, n, which is 6. Thus, the molecular formula is C†H†, which simplifies to CH, since both subscripts can be divided by 3.
Therefore, the correct molecular formula of the compound CH is [tex]C_6H_6[/tex].
"What does this observation tell us about the enthalpy of C2H5Cl(g) as compared with the enthalpy of C2H5Cl(l)
Answer:
Enthalpy of gas is greater than that of liquid
Explanation
Ethyl chloride (C2H5Cl) boils at 12 degrees Celsius. When liquid C2H5Cl under pressure is sprayed on a room-temperature (25 degrees Celsius) surface in the air, the surface is cooled considerably.
this preamble could precede the question
Enthalpy
it can defined as the ability of a substance to change at constant pressure, enthalpy tells how much heat and work was added or removed from the substance. Enthalpy is similar to energy, but not the same. When a substance grows or shrinks, energy is used up or released.
The total heat content of a system is known as Enthalp.it is a thermodynamic property. It is the internal energy of the system plus the product of pressure and volume.
H=U+PV........................1
A student performs an experiment to determine the density of a sugar solution. She obtains the following results: 1.11 g/mL, 1.81 g/mL, 1.95 g/mL, 1.75 g/mL. If the actual value for the density of the sugar solution is 1.75 g/mL, which statement below best describes her results
A) Her results are precise, but not accurate.
B) Her results are accurate, but not precise.
C) Her results are both precise and accurate
D) Her results are neither precise nor accurate.
E) It isn't possible to determine with the information given.
The student's results are best described as accurate but not precise, as one measurement matches the actual value, but the other measurements are inconsistent with each other and the actual value.
You are trying to determine the density of a sugar solution and have obtained results of 1.11 g/mL, 1.81 g/mL, 1.95 g/mL, and 1.75 g/mL, with the actual density being 1.75 g/mL.
When comparing these results to the actual value, we can see that one of the results matches the actual value, indicating accuracy for that particular measurement.
However, the other measurements are quite different from the actual value and from each other, indicating a lack of precision. Precision refers to how close multiple measurements are to each other, regardless of whether they are close to the actual value (which is accuracy).
So, the statement that best describes the results is:
(A) Her results are precise, but not accurate.(B) Her results are accurate, but not precise.(C) Her results are both precise and accurate.(D) Her results are neither precise nor accurate.(E) It isn't possible to determine with the information given.The most appropriate choice is (B), Her results are accurate, but not precise, because one measurement matches the actual value while the other measurements do not consistently match each other.
Draw a structure for an amine that exhibits a molecular ion at M+ = 87 and that produces fragments at m/z = 72 and m/z = 58.
Answer:
The amine is N-methylbutan-2-amine and the structure is drawn below.
Explanation:
In order to find out the correct structure for said data we will first try out to find the molecular formula by applying rule of thirteen.
Rule of Thirteen:
First divide the parent peak value by 13 as,
= 87 ÷ 13
= 6.6923
Now, multiply 13 by 6,
= 13 × 6 (here, 6 specifies number of carbon atoms)
= 78
Now subtract 78 from 87,
= 87 - 78
= 9
Add 9 into 11,
= 9 + 6
= 15 (hydrogen atoms)
So, the rough formula we have is,
C₆H₁₅
Now, add one Nitrogen atom to above formula and subtract one Carbon and 2 Hydrogen atoms as these numbers are equal to atomic mass of Nitrogen atom as,
C₆H₁₅ -------N--------> C₅H₁₃N
Hence, the molecular formula is C₅H₁₃N.
The correct structural formula is being predicted by using the stability of the fragments formed by the cleavage of the parent molecule. Therefore, as shown below, following secondary (stable) carbocations are formed as the major fragments.
Two different compounds are obtained by combining nitrogen with oxygen. The first compound results from combining 46.7 gg of NN with 53.3 gg of OO, and the second compound results from combining 17.9 gg of NN and 82.0 gg of OO. Calculate the ratio of the mass ratio of OO to NN in the second compound to the mass ratio of OO to NN in the first compound.
Answer:
4.02
Explanation:
The mass ratios will be given by dividing the mass of O₂ into the mass of N₂.
So lets do our calculations:
First Compound:
53.3 g O₂ / 46.7 g N₂ = 1.14
Second Compound:
82.0 g O₂ / 17.9 g N₂ = 4.58
Ratio = 4.58 / 1.14 = 4.02
This result for all practical purposes is a whole number, and it is telling us that there are 4 times as many oxygen atoms in the second coumpound as in the first compound. This is so because the ratio we just calculated is also the ratio in mol atoms:
Ratio = [ mass O₂ / MW O2/ mass N₂/ MW N₂] 2nd compound / [mass O₂ / MW O2/ mass N₂/ MW N₂ !st compound]
and the molecular weights cancel each other.
The only N and O compounds that follow this ratio are N₂O₄ and N₂O, and this question could be made in a multiple choice to match formulas.
Answer: 2
Explanation:
Two different compounds are obtained by combining nitrogen with oxygen. The first compound results from combining 46.7 g of N with 53.3 g of O, and the second compound results from combining 30.4 g of N and 69.6 g of O. Calculate the ratio of the mass ratio of O to N in the second compound to the mass ratio of O to N in the first compound.
2
The ratio of the mass ratio of O to N in the second compound to the mass ratio of O to N in the first compound is calculated as
mass of Omass of N in second compound ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯mass of Omass of N in first compound = 2.291.14 = 2
Thus, the mass ratio of O to N in the second compound is two times the mass ratio of O to N in the first compound.
Neon has three naturally occuring isotopes. The abundance of 20Ne is 90.48% and 22Ne is 9.25%. What is the percent abundance of 21Ne?
Answer: The percent abundance of Ne-21 isotope is 0.27 %
Explanation:
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex] .....(1)
Let the fractional abundance of Ne-21 isotope be x
For isotope 1 (Ne-20) :Mass of isotope 1 = 20 amu
Percentage abundance of isotope 1 = 90.48 %
Fractional abundance of isotope 1 = 0.9048
For isotope 2 (Ne-21) :Mass of isotope 2 = 21 amu
Fractional abundance of isotope 2 = x
For isotope 3 (Ne-22) :Mass of isotope 3 = 22 amu
Percentage abundance of isotope 3 = 9.25 %
Fractional abundance of isotope 3 = 0.0925
Average atomic mass of neon = 20.18 amu
Putting values in equation 1, we get:
[tex]20.18=[(20\times 0.9048)+(21\times x)+(22\times 0.0925)][/tex]
x = 0.0027
Percentage abundance of Ne-21 isotope = [tex](0.0027\times 100)=0.27\%[/tex]
Hence, the percent abundance of Ne-21 isotope is 0.27 %
The abundance of the isotope 21Ne of Neon can be found by subtracting the sum of the abundances of the other two isotopes, 20Ne and 22Ne, from 100%. This calculation yields an abundance of approximately 0.27% for 21Ne.
Explanation:Neon has three naturally occurring isotopes: 20Ne, 21Ne, and 22Ne. Given the percentage abundances of 20Ne and 22Ne, we can find the abundance of 21Ne by subtracting the sum of these percentages from 100%. Specifically,
Abundance of 21Ne = 100% - (90.48% + 9.25%)
So, the abundance of 21Ne is approximately 0.27%.
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Determine if the results of the following word problem adhere to the Law of Conservation of Mass. 1. A chemist combines 4.9 g of hydrogen gas with 9.4 grams of nitrogen gas to form 11.4 g of ammonia. 2.9 g of nitrogen gas is remaining
Answer: The results agree with the law of conservation of mass
Explanation:
The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. On the reactant side, the total mass of reactants is 14.3g and the total product masses is also 14.3g. That implies that no mass was !most in the reaction. The sum of masses on the left hand side corresponds with sum of masses on the right hand side of the reaction equation.
Consider the chemical reaction :
????????H4 + ????????2 → c????????2 + ????H2????,
where ????, ????, c, and ???? are unknown positive integers. The reaction must be balanced; that is, the number of atoms of each element must be the same before and after the reaction. For example, because the number of oxygen atoms must remain the same, 2????=2c+????. While there are many possible choices for ????, ????, c, and ???? that balance the reaction, it is customary to use the smallest possible integers.
Balance this reaction.
The question is incomplete, here is the complete question:
Consider the chemical reaction:
[tex]aCH_4+bO_2\rightarrow cCO_2+dH_2O[/tex]
where a, b, c, and d are unknown positive integers. The reaction must be balanced; that is, the number of atoms of each element must be the same before and after the reaction. For example, because the number of oxygen atoms must remain the same, 2b=2c+d. While there are many possible choices for a, b, c, and d that balance the reaction, it is customary to use the smallest possible integers.
Balance this reaction.
Answer: The value of a, b, c and d in the given chemical equation are 1, 2, 1 and 2 respectively.
Explanation:
Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.
This also means that total mass on the reactant side must be equal to the total mass on the product side.
For the given chemical reaction:
[tex]aCH_4+bO_2\rightarrow cCO_2+dH_2O[/tex]
On reactant side:
Number of carbon atoms = 1
Number of hydrogen atoms = 4
Number of oxygen atoms = 2
On product side:
Number of carbon atoms = 1
Number of hydrogen atoms = 2
Number of oxygen atoms = 3
We need to balance the number of hydrogen and oxygen atoms. So, the balanced equation for the given reaction becomes:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
Hence, the value of a, b, c and d in the given chemical equation are 1, 2, 1 and 2 respectively.
Draw the molecular structures of citrate, isocitrate, α-ketoglutarate, and Succinyl-CoA in a pathway sequence. Show which carbons came from acetyl-CoA and which go off as CO2.
Answer: check the picture
Explanation:
Answer:
In the oxidation of Pyruvate to Acetyl CoA one carbon atom is released as CO2, However, the oxidation of the remaining two carbon atoms in acetate to CO2, requires a complex, eight step pathway, the citric-acid cycle.
Explanation:
Overall, one turn of the citric acid cycle releases two carbon dioxide molecules and produces three {NADH} one {FADH2} and one ATP or GTP. The Citric acid cycle goes around twice for each molecule of glucose that enters cellular respiration because there are two pyruvate and thus, two acetyl {CoA}s are made per glucose.
26.6 mL of 2.50 M stock solution of sucrose is diluted to 50.0 mL. A 16.0 mL sample of the resulting solution is then diluted to 45.0 mL. A 20.0 mL sample of this solution is then diluted to 75.0 mL. What is the concentration of sucrose in the final solution?
Answer:
In the final solution, the concentration of sucrose is 0.126 M
Explanation:
Hi there!
The number of moles of solute in the volume taken from the more concentrated solution will be equal to the number of moles of solute in the diluted solution. Then, the concentration of the first solution can be calculated using the following equation:
Ci · Vi = Cf · Vf
Where:
Ci = concentration of the original solution
Vi = volume of the solution taken to prepare the more diluted solution.
Cf = concentration of the more diluted solution.
Vf = volume of the more diluted solution.
For the first dillution:
26.6 ml · 2.50 M = 50.0 ml · Cf
Cf = 26.6 ml · 2.50 M / 50.0 ml
Cf = 1.33 M
For the second dilution:
16.0 ml · 1.33 M = 45.0 ml · Cf
Cf = 16.0 ml · 1.33 M / 45.0 ml
Cf = 0.473 M
For the third dilution:
20.0 ml · 0.473 M = 75.0 ml · Cf
Cf = 20.0 ml · 0.473 M / 75.0 ml
Cf = 0.126 M
In the final solution, the concentration of sucrose is 0.126 M
Which compound (s) has(have) a molecular ion at 117 and a peak at 1720 cm-1 in its IR spectrum?
Answer:
Ethyl Acetate with a chemical configuration of : CH3CO2CH2CH3
Explanation:
Ethyl Acetate has a density of 902 kg/m³, molar mass of 88.11 g/mol and posses a boiling point of 77.1 °C.
It is to be understood that ethyl acetate is the ester of ethanol and it is mostly produced in mass production and most especially used in the production of domestic materials especially materials like glue, efostics and dissolving agent. This is a highly toxic and flammable substance. However colorless and possess a sweet smell, it can be highly poisonous when ingested.
The following reaction between ethyl alcohol and ethyl bromide forms diethyl ether via a substitution reaction (followed by deprotonation).cannot explain figure
Answer:
see explanation below
Explanation:
The question is incomplete. The missing parts are a) determine the electrophylic site. b) determine the nucleophylic site.
In order to do this, we need to write the reaction and do the mechanism. The nucleophylic site will be the site where the nucleophyle attacks to form the product. In this case the site is the carbon next to the bromine. In this place the Oxigen which is the nucleophyle goes. The electrophyle is the site where one atom substract to complete it's charges. In this case, the electrophyle is usually the hydrogen, so the site will be next to the oxygen after the nucleophyle attack.
You can see it better in the attached picture.
The reaction between ethyl alcohol and ethyl bromide forms diethyl ether via a substitution reaction, where a hydrogen atom of the alcohol is replaced with ethyl from the bromide. After this, deprotonation occurs to remove excess hydrogen and create the stable diethyl ether.
Explanation:The question refers to the reaction between ethyl alcohol and ethyl bromide to produce diethyl ether, a type of ether. This reaction happens via a substitution reaction, during which a hydrogen atom of the ethyl alcohol is replaced by an ethyl group from the ethyl bromide. After the substitution, a deprotonation process occurs to form the final product, diethyl ether.
In chemical terms, the substitution reaction occurs when an alcohol, like ethyl alcohol, reacts with a halogenoalkane, like ethyl bromide, in the presence of sulphuric acid (H2SO4). The result is the formation of an ether, in this case, diethyl ether. A deprotonation process takes place to remove an excess hydrogen proton to form the final stable product.
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An example of metal-bearing deposit formed when a sulphide liquid separated from an ultramafic magma by becoming supersaturated and immiscible is
Answer: SUDBURY, ONTARIO.
Explanation: SUDBURY,is a place in Ontario, Canada known to have been formed as a result of the separation of Sulphur liquid from ultramafic Magma by becoming supersaturated and immiscible. Over 100millions tons of Sulphur liquid and some quantities of other metals like Nickel,iron have been released into the atmosphere as a result of mining activities of that period.
Sudbury basin is famous for its geological feature that hosts about the largest concentrations of nickel-copper sulphides in the world is believed to have been formed by a meteorite impact 1.8 billion years ago.