A solution is prepared by adding 1.43 mol of kcl to 889 g of water. the concentration of kcl is ________ molal. a solution is prepared by adding 1.43 mol of kcl to 889 g of water. the concentration of kcl is ________ molal. 1.27 × 103 1.61 1.61 × 10-3 0.622 622
The concentration of the KCl solution is 1.61 molal.
Explanation:The concentration of a solution can be described by its molality (m), which is the number of moles of solute per kilogram of solvent. To find the molality of the KCl solution, we need to calculate the number of moles of KCl and the amount of water in kilograms. Given that 1.43 mol of KCl is dissolved in 889 g of water, we convert the mass of water to kilograms (889 g = 0.889 kg). Now we can calculate the molality:
Molality (m) = moles of KCl / kilograms of water
Molality (m) = 1.43 mol / 0.889 kg = 1.61 molal
How many atoms are in a single molecule of water H2O
There are 3 atoms in total. Two come from Hydrogen hence the H2. The third comes from the single Oxygen (O). Hope this helps!
Three atoms make up the water molecule (H2O) as a whole.
What is a molecule?According to the context, the term may or may not include ions that meet this requirement. A molecule is a collection of two or more atoms held together by attractive forces known as chemical bonds.
The same atoms can combine in various ratios to create various molecules. For instance, water (H2O) is made up of two hydrogen atoms and one oxygen atom, whereas hydrogen peroxide (H2O2) is made up of two hydrogen atoms and two oxygen atoms.
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Three beakers contain clear, colorless liquids. one beaker contains pure water, another contains salt water, and another contains sugar water. how can you tell which beaker is which? (
Final answer:
To distinguish between pure water, salt water, and sugar water, conduct tests such as adding silver nitrate to detect chloride ions or observe evaporation rates in a sealed chamber. Salt water may form a precipitate upon adding silver nitrate, while pure water has a higher vapor pressure than sugar water.
Explanation:
To distinguish between a beaker of pure water, salt water, and sugar water, we need to conduct simple experiments that reveal the properties of each liquid as they differ in terms of their chemical components and physical properties.
Adding silver nitrate to a clear colorless solution that forms a pale yellow precipitate indicates the presence of chloride ions, suggesting a salt water solution. A beaker containing pure water would not react, while sugar water would also not produce a precipitate.
When considering evaporation in a sealed chamber, the beaker with pure water will have a higher vapor pressure compared to the sugar water due to the presence of sugar molecules at the surface, which inhibit the evaporation rate. Over time, all of the water from the beaker with higher vapor pressure (pure water) will evaporate and condense in the beaker with the sugar solution.
To differentiate between the three beakers, we must observe the reactions mentioned above (if testing chemically) or deduce based on the properties like vapor pressure and the effects they have on evaporation and condensation processes.
Which of the following is the last step in performing a titration
1) finding out which pH indicatorworks
2) determining the concentration of an unknown base
3) finding the number of moles of a product produced in a reaction
4) determining the molecular masses of the products in the reaction
Answer: determining the concentration of an unknown base
Explanation: The last step in performing a titration will lead to the determination of the concentration of the unknown base or an unknown quantity.
The other steps like finding out which pH indicator works, finding the number of moles of a product produced in a reaction and determining the molecular masses of the products in the reaction comes in between the procedure.
At the start of a measurement, a radioisotope has 10,000 unstable nuclei. over 2 days 5,000 of these unstable nuclei undergo radioactive decay to stable nuclei. what is the half life of this radioisotope?
Based on the experimental results, which solvent provided the second best separation of the analyte?
a. 0.5% nacl solution
b. vinegar
c. 0.2% nacl solution
d. all solvents had the same degree of separation
e. distilled water f. 70% isopropyl alcohol
Calculate the molality of a solution formed by dissolving 34.8 g of lii in 500.0 ml of water. calculate the molality of a solution formed by dissolving 34.8 g of lii in 500.0 ml of water. 0.520 m 0.260 m 0.254 m 0.696 m 0.130 m
The molality of solution formed by dissolving 34.8 g of LiI in 500 mL of water is [tex]\boxed{0.520{\text{ m}}}[/tex].
Further Explanation:
Concentration terms are used to determine concentration of various components present in any mixture. Some of these are mentioned below.
1. Molarity (M)
2. Mole fraction (X)
3. Molality (m)
4. Parts per million (ppm)
5. Mass percent ((w/w) %)
6. Volume percent ((v/v) %)
7. Parts per billion (ppb)
Molality is defined as moles of solute that are present per kilograms of solvent. It is represented by m and its unit is mol/kg. The expression for molality of solution is as follows:
[tex]{\text{Molarity of solution}} = \dfrac{{{\text{Moles }}\left( {{\text{mol}}} \right){\text{of solute}}}}{{{\text{Mass }}\left( {{\text{kg}}} \right){\text{ of solvent}}}}[/tex]
Given information:
Mass of LiI: 34.8 g
Volume of water: 500.0 mL
To calculate:
Molality of LiI solution
How to proceed:
Step1: Moles of LiI that are present in 34.8 g of LiI has to be determined. This is done with the help of equation (1).
The formula to calculate the moles of LiI is as follows:
[tex]{\text{Moles of LiI}} = \dfrac{{{\text{Mass of LiI}}}}{{{\text{Molar mass of LiI}}}}[/tex] …… (1)
The mass of LiI is 34.8 g.
The molar mass of LiI is 133.841 g/mol.
Substitute these values in equation (1).
[tex]\begin{aligned}{\text{Moles of LiI}} &= \frac{{{\text{34}}{\text{.8 g}}}}{{{\text{133}}{\text{.841 g/mol}}}} \\&= 0.26{\text{ mol}} \\\end{aligned}[/tex]
Step 2: Mass of water is to be evaluated.
At standard temperature and pressure conditions, one liter of water can be considered equivalent to 1 kilograms of water.
Therefore mass of water can be calculated as follows:
[tex]\begin{aligned}{\text{Mass of water}} &= \left( {500.0{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right)\left( {\frac{{1{\text{ kg}}}}{{1{\text{ L}}}}} \right) \\&= 0.5{\text{ kg}} \\\end{aligned}[/tex]
Step 3: Molality of LiI solution is to be calculated. This is done with the help of equation (2).
The formula to calculate molality of LiI solution is as follows:
[tex]{\text{Molality of LiI solution}} = \dfrac{{{\text{Moles of LiI}}}}{{{\text{Mass of water}}}}[/tex] …… (2)
Substitute 0.26mol for moles of LiI and 0.5 kg for mass of water in equation (2).
[tex]\begin{aligned}{\text{Molality of LiI solution}} &= \frac{{{\text{0}}{\text{.26 mol}}}}{{{\text{0}}{\text{.5 kg}}}} \\&= 0.520{\text{ m}} \\\end{aligned}[/tex]
Hence, molality of given solution comes out to be 0.520 m.
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Concentration terms
Keywords: concentration, concentration terms, solutions, molarity, molality, LiI, moles, mass, molar mass, 0.520 m, 0.5 kg, 0.26 mol.
Consider the formation of ammonia in two experiments. (a) to a 1.00-l container at 727°c, 1.30 mol of n2 and 1.65 mol of h2 are added. at equilibrium, 0.100 mol of nh3 is present. calculate the equilibrium concentrations of n2 and h2 for the reaction: 2 nh3(g) n2(g) + 3 h2(g). n2 1.25 mol/l h2 1.50 mol/l find kc for the reaction. kc = 422 m2 (b) in a different 1.00-l container at the same temperature, equilibrium is established with 8.34 ✕ 10−2 mol of nh3, 1.50 mol of n2, and 1.25 mol of h2 present. calculate kc for the reaction: nh3(g) 1 2 n2(g) + 3 2 h2(g). 20.5 m (c) what is the relationship between the kc values in parts (a) and (b)? why aren't these values the same?
In the formation of ammonia, different initial amounts of reactants result in different Kc values at equilibrium. While the temperature is constant, varying concentrations of reactants and products lead to different equilibrium constants Kc.
Explanation:The chemical equation is given by 2NH3(g) = N2(g) + 3H2(g). In the first scenario, you start with 1.30 mol of N2 and 1.65 mol of H2 in a 1.00 L container, with 0.100 mol of NH3 at equilibrium. This implies a decrease in N2 and H2 by x amount and an increase in NH3 by 2x. Given the stoichiometric relationships, we can calculate that the equilibrium concentrations of N2 and H2 are 1.20 M and 1.30 M respectively.
Subsequently, we can calculate Kc using the equilibrium concentrations, Kc = [NH3]^2/(N2][H2]^3) = (0.10)^2/(1.20 * (1.55)^3) = 422 M^-2.
In the second scenario, the same calculations result in a Kc = 20.5 M^-2. The discrepancy in the Kc values arises due to the differences in the initial amounts of reactants and the products formed. Although the temperature is constant and the reaction is the same, the amount of reactants used and the products formed vary which impacts the Kc value.
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How much heat is required to raise the temperature of 475 grams of solid aluminum from room temperature (295k) to liquid aluminum at 1050k?
528.2 kJ heat is required to raise the temperature of 475 grams of solid aluminum from room temperature (295k) to liquid aluminum at 1050k.
What is heat ?Heat is the result of the movement of kinetic energy within a material or an item, or from an energy source to a material or an object. Radiation, conduction, and convection are the three mechanisms through which such energy can be transferred.
Heat is the quantity of energy that is transferred from one structure to its surrounds as a result of a change in temperature. Heat is the transfer of kinetic energy between one media or item and another, or from an energy source to a medium or object.
The total amount of heat required
∑q = q1 + q2 + q3
q1 = mCΔT
= 475 g x 0.902 J/g-oC x ( 933.47 - 295 ) K
= 273,552.47J
= 273 kJ
ΔT = ( 933.47 - 295 ) K
= 638.47 K
ΔT = ( 933.47 K -273) - (295 K - 273)
= 638.47°C
For ΔT temperature in K is same as in °C
q2 = mol ΔHf
mol = 475 g x 1 mol /27g
= 17.6 mol
q2 = mol ΔHf = 17.6 mol x 10.79 kJ/mol
= 189.9 kJ
q3 = mCΔT
= 0.475 kg x 1.18 kJ/kg-oC (1050 - 933.47)K
= 65.3 kJ
∑q = q1 + q2 + q3
= 273 + 189.9 + 65.3
= 528.2 kJ
Thus,528.2 kJ heat is required to raise the temperature of 475 grams of solid aluminum from room temperature (295k) to liquid aluminum at 1050k.
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What is the theoretical yield of ammonia, in kilograms, that we can synthesize from 5.22 kg of h2 and 31.5 kg of n2?
Given:
Mass of H2 = 5.22 kg = 5220 g
Mass of N2 = 31.5 kg = 31500 g
To determine:
Theoretical yield of NH3
Explanation:
The balanced chemical reaction is:
N2 + 3H2 → 2NH3
1 mole of N2 combines with 3 moles of H2 to form 2 moles of NH3
# moles of N2 = 31500 g/ 28 g.mol-1 = 1125 moles
# moles of H2 = 5200 g/ 1 g.mol-1 = 5200 moles
Therefore N2 is the limiting reagent
Based on the stoichiometry:
1 mole of N2 forms 2 moles of NH3
Thus, 1125 moles of N2 will yield : 1125 * 2 = 2250 moles of NH3
Mass of NH3 = 2250 moles * 17 g/mole = 38250 g = 38.3 kg
Ans: Theoretical yield of NH3 = 38.3 kg
Answer : The theoretical yield of ammonia is, 29.58 Kg
Solution : Given,
Mass of [tex]H_2[/tex] = 5.22 Kg = 5220 g
Mass of [tex]N_2[/tex]= 31.5 Kg = 31500 g
Molar mass of [tex]H_2[/tex] = 2 g/mole
Molar mass of [tex]N_2[/tex] = 28 g/mole
Molar mass of [tex]NH_3[/tex] = 17 g/mole
First we have to calculate the moles of [tex]H_2[/tex] and [tex]N_2[/tex] .
Moles of [tex]H_2[/tex]= [tex]\frac{\text{ given mass of }H_2}{\text{ molar mass of }H_2}= \frac{5220g}{2g/mole}=2610moles[/tex]
Moles of [tex]N_2[/tex] = [tex]\frac{\text{ given mass of }N_2}{\text{ molar mass of }N_2}= \frac{31500g}{28g/mole}=1125moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction will be,
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
From the balanced reaction we conclude that
1 mole of [tex]N_2[/tex] react with 3 moles of [tex]H_2[/tex]
1125 moles of [tex]N_2[/tex] react with [tex]3\times 1125=3375[/tex] moles of [tex]H_2[/tex]
That means [tex]H_2[/tex] is a limiting reagent and [tex]N_2[/tex] is an excess reagent.
Now we have to calculate the moles of ammonia.
From the reaction we conclude that,
3 moles of [tex]H_2[/tex] react to give 2 moles of ammonia
2610 moles of [tex]H_2[/tex] react to give [tex]\frac{2}{3}\times 2610=1740[/tex] moles of ammonia
Now we have to calculate the mass of ammonia.
[tex]\text{Mass of }NH_3=\text{Moles of }NH_3\times \text{Molar mass of }NH_3[/tex]
[tex]\text{Mass of }NH_3=(1740moles)\times (17g/mole)=29580g=\frac{29580}{1000}=29.58Kg[/tex]
Therefore, the theoretical yield of ammonia is, 29.58 Kg
Write a balanced chemical equation for the reaction. feso4 and pb(no3)2 express your answer as a chemical equation. identify all of the phases in your answer.
Answer: The balanced chemical equation is written above.
Explanation:
Balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side will be equal to the total number of individual atoms on the product side. These equations follow law of conservation of mass.
When ferrous sulfate reacts with lead (II) nitrate, leads to the production of iron (II) nitrate and lead (II) sulfate.
The chemical equation for the above reaction follows:
[tex]FeSO_4(aq.)+Pb(NO_3)_2(aq.)\rightarrow Fe(NO_3)_2(aq.)+PbSO_4(s)[/tex]
By Stoichiometry of the reaction:
1 mole of aqueous solution of ferrous sulfate reacts with 1 mole of aqueous solution of lead (II) nitrate to produce 1 mole of aqueous solution of iron (II) nitrate and 1 mole of solid lead (II) sulfate.
Hence, the balanced chemical equation is written above.
Find the molality of the solution if 42 grams of lithium chloride (LiCl) are dissolved in 3.6 kg of water.
0.275 m
11.7 m
0.990 m
0.857 m
Final answer:
To calculate the molality of the solution, divide the number of moles of lithium chloride (0.990 moles) by the mass of the water in kilograms (3.6 kg), resulting in a molality of 0.275 m.
Explanation:
The molality of a solution is calculated by finding the number of moles of solute and dividing by the mass of the solvent in kilograms. To find the molality of the solution with 42 grams of lithium chloride (LiCl) dissolved in 3.6 kg of water, we first convert grams to moles using the molar mass of LiCl.
The molar mass of LiCl is 42.39 g/mol. So, if we divide 42 grams by the molar mass, we get the number of moles of LiCl:
Moles of LiCl = 42 g / 42.39 g/mol = 0.990 moles of LiCl
We then use this number of moles and the mass of the solvent in kilograms to find the molality:
Molality (m) = Moles of solute / Mass of solvent (kg) = 0.990 moles / 3.6 kg = 0.275 m
Therefore, the molality of the solution is 0.275 m.
How many grams are in 4.00 mol of sodium chloride (nacl)?
Mole measure the number of elementary entities of a given substance that are present in a given sample. The mass of 4.00 mole of sodium chloride is 233.76g.
What is mole?The SI unit of amount of substance in chemistry is mole. The mole is used to measure the quantity of amount of substance. One mole of any element contains 6.022×10²³ atoms which is also called Avogadro number.
Mathematically,
mole of sodium chloride = mass ÷ Molar mass of sodium chloride
number of mole of sodium chloride=4mol
Molar mass sodium chloride =58.44 g/mole
Substituting the given values in the above equation, we get
4mole of sodium chloride=mass mass of sodium chloride÷ 58.44 g/mole
mass of sodium chloride = 4× 58.44 g/mole
mass of sodium chloride= 233.76g
Therefore the mass of 4.00 mole of sodium chloride is 233.76g.
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Which is a Not scientific question?
What is the pH of a solution of potassium hydroxide?
Which base is the most exciting to study?
Will HF neutralize a solution of HCl?
Does adding sodium chloride change the pH of a solution?
An old sample of concentrated sulfuric acid to be used in the laboratory is approximately 98.1 percent h2so4 by mass. calculate the molality and molarity of the acid solution. the density of the solution is 1.83 g/ml.
The molality of a 98.1% sulfuric acid solution is calculated to be 526.32 mol/kg and the molarity is found to be 18.32 M.
Explanation:To find the molality and molarity of sulfuric acid in a solution, we need to know the mass and volume of the solution. Given that the solution is 98.1% sulfuric acid by mass, we can say that for every 1000g (or 1kg) of solution, there are 981g of sulfuric acid. The molecular weight of sulfuric acid is 98.08 g/mol, so we can convert this to moles to find the molality.
Molality = moles of solute / mass of solvent in kg
The moles of sulfuric acid = 981g / 98.08 g/mol = 10 mol. The mass of the water in the solution is 1000g - 981g = 19g, or 0.019kg. So, the molality of the solution is 10 mol / 0.019 kg = 526.32 mol/kg.
To find the molarity, we need to know the volume of the solution. Given the density of the solution is 1.83 g/ml, we can say that 1kg of solution is 1000g / 1.83g/ml = 546ml = 0.546L. So, the molarity = moles of solute / volume of solution in liters = 10 mol / 0.546 L = 18.32 M.
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Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 35.9°c.
The root mean square velocity (RMS velocity) is a method to find a single velocity value for the particles. The average velocity of gas particles in a mixture is found using the root mean square velocity formula:
μrms = sqrt (3RT/M)
Where,
R = universal gas constant = 8.3145 (kg·m^2/s^2)/K·mol
T = temperature = 35.9 ˚C = 309.05 K
M = molar mass = 28 * 10^-3 kg / mol
Substituting the given values into the equation:
μrms = sqrt [(3 * 8.3145 (kg·m^2/s^2) /K·mol) * (309.05 K) / (28 * 10^-3 kg / mol))]
μrms = sqrt (275,313.8813 m^2)
μrms = 524.7 m / s (ANSWER)
A newly discovered element has two isotopes. one has an atomic weight of 120.9038 amu with 57.25% abundance. the other has an atomic weight of 122.8831 amu. what is the atomic weight of the element?
the atomic weight of the element : 122.22 amu
Further explanationThe elements in nature have several types of isotopes
Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.
So Isotopes are elements that have the same Atomic Number (Proton)
Atomic mass is the average atomic mass of all its isotopes
In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu
So the atomic mass obtained is the mass of the atom relative to the 12th carbon atom
Mass atom X = mass isotope 1 . % + mass isotope 2.%
An atomic mass unit = amu is a relative atomic mass of 1/12 the mass of an atom of carbon-12.
The 'amu' unit has now been replaced with a unit of 'u' only
for example, Carbon has 3 isotopes, namely ₆¹²C, ₆¹³C, and ₆¹⁴C
Element has two isotopes. 120.9038 amu with 57.25% abundance and the other has an atomic weight of 122.8831 amu
Mass atom element X = mass isotope 1 . % + mass isotope 2.%
Mass atom element X = 120.9038 . 57.25% + 122.8831. 42.75%
Mass atom element X = 69.22 + 52.53 = 122.22 amu
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The world's oceans contain approximately 7.6 x 1019 moles of water molecules. Based on this, how many water molecules are there in the world's oceans?
Calculate the ph of a 0.10 m solution of barium hydroxide, ba(oh)2. express your answer numerically using two decimal places.
equation is
pH=14+log(.2)
Why does potassium permanganate diffuse faster than methylene blue?
Diffusion of a substance is affected by its molecular weight and mobility of molecules. Potassium permanganate diffuse faster than methylene blue because, it is lighter than methylene blue.
What is diffusion ?Diffusion is the process of movement of molecules from from higher concentration area to that of lower concentration. The movement depends on the molecular weight, its density and mobility of its ions.
Potassium permanganate molecules are smaller and have lesser weight (150 g/mol) than methylene blue. Methylene blue is having higher molecular weight (319.85 g/mol) since it contains big aromatic rings.
Therefore, potassium permanganate diffuse faster than methylene blue.
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Tetrahydrocannabinol (thc) is composed of 80.24% carbon, 9.62% hydrogen and 10.14% oxygen by mass. what is the empirical formula of thc?
One mole of an ideal gas is sealed in a 22.4-l container at a pressure of 1 atm and a temperature of 273 k. the temperature is then increased to 311 k , but the container does not expand. what will the new pressure be? part a the most appropriate formula for solving this problem includes only which variables? enter the required variables, separated by commas (e.g., p,v,t).
The appropriate formula for solving this problem includes the variables pressure and temperature (P, T).
Explanation:To solve this problem, we can use the combined gas law formula, which includes the variables pressure (P), volume (V), and temperature (T).
The combined gas law formula is as follows: P1V1/T1 = P2V2/T2
In this problem, the volume is constant, so we only need to consider pressure and temperature. Therefore, the appropriate formula for solving this problem includes the variables pressure and temperature (P, T).
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1. After calculation, the new pressure of the gas is approximately 1.139 atm. 2.The variables involved are pressure (P) and temperature (T).
1. To determine the new pressure of the gas after the temperature increase while keeping the volume constant, we should use the combined gas law, which states:
P₁/T₁ = P₂/T₂
Given:
Initial pressure (P₁) = 1 atmInitial temperature (T₁) = 273 KFinal temperature (T₂) = 311 KWe need to find the final pressure (P₂)Rearrange the formula to solve for P₂:
[tex]\[P_2 = P_1 \cdot \frac{T_2}{T_1}\][/tex]
Substitute the given values:
[tex]P_2 = 1 \, \text{atm} \cdot \left( \frac{311 \, \text{K}}{273 \, \text{K}} \right)[/tex]
Calculate:
P₂ ≈ 1.139 atm
Therefore, the new pressure will be approximately 1.139 atm.
2. The most appropriate formula for solving this problem includes only the following variables: P, T.
In the reaction of 1.15 g of salicylic acid, molar mass 138.12 g/mol, with 2.83 g of acetic anhydride, molar mass 102.10 g/mol, a student obtained 1.21 g of acetylsalicylic acid, molar mass 180.17 g/mol. What is the percent yield?
The
balanced reaction would be written as:
C7H6O3 + C4H6O3--->C9H8O4 + HC2H3O2
To determine the percent yield, we need to first
determine the theoretical yield if the reaction were to proceed completely.
Then, we divide the actual yield that is given to the theoretical yield times
100 percent. The limiting reactant from the reaction would be salicylic acid.
We do as follows:
Theoretical yield: 1.15 g C7H6O3 ( 1 mol /
138.12 g ) ( 1 mol C9H8O4 / 1 mol C7H6O3 ) ( 180.17 g / mol ) = 1.50 g C9H8O4 should be produced
Percent yield = 1.21 / 1.50 x 100 = 80.66 %
Thus, only 80.66% of the
theoretical C9H8O4 is being produced.
The percent yield of the reaction is 14549%.
Explanation:To find the percent yield, we need to compare the actual yield (1.21 g of acetylsalicylic acid) to the theoretical yield, which is the maximum amount of acetylsalicylic acid that could be produced based on the reactants and their stoichiometry.
We can calculate the theoretical yield using the molar ratios of the reactants. The balanced equation for the reaction is:
C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2
From the equation, we can see that 1 mole of salicylic acid reacts with 1 mole of acetic anhydride to produce 1 mole of acetylsalicylic acid. Therefore, we can use the moles of salicylic acid and acetic anhydride to calculate the theoretical yield of acetylsalicylic acid.
The moles of salicylic acid can be calculated using its mass and molar mass:
moles of salicylic acid = mass of salicylic acid / molar mass of salicylic acid = 1.15 g / 138.12 g/mol = 0.00833 mol
The moles of acetic anhydride can be calculated using its mass and molar mass:
moles of acetic anhydride = mass of acetic anhydride / molar mass of acetic anhydride = 2.83 g / 102.10 g/mol = 0.0278 mol
Since the balanced equation shows a 1:1 molar ratio between salicylic acid and acetylsalicylic acid, the moles of salicylic acid and acetic anhydride are equal to the moles of acetylsalicylic acid produced.
Therefore, the theoretical yield of acetylsalicylic acid is 0.00833 mol.
The percent yield can be calculated using the formula:
percent yield = (actual yield / theoretical yield) * 100
Substituting the values, we get:
percent yield = (1.21 g / 0.00833 mol) * 100 = 14549%
The percent yield is 14549%.
Find the molality of the solution if 2.5 moles of CaBr2 are dissolved in 5 kg of water.
-500 m
-5 x 10-4 m
-0.5 m
-2 m
Given the formula c1v1=c2v2, where c indicates concentration and v indicates volume, which equation represents the correct way to find the concentration of the dilute solution (c2)
What property of liquid oxygen makes it especially difficult and potentially harmful to work with at home?
Calculate the ph of a 1.60 mch3nh3cl solution. kb for methylamine, ch3nh2, is 3.4×10−4m. calculate the of a 1.60 solution. for methylamine, , is . 12.39 8.82 1.61 5.18
The base dissociation constant or Kb is a value used to measure the strength of a specific base in solution. To determine the percent ionization of the substance we make use of the Kb given. Methylamine or CH3NH2 when in solution would form ions:
CH3NH2 + H2O < = > CH3NH3+ + OH-
Kb is expressed as follows:
Kb = [OH-] [CH3NH3+] / [CH3NH2]
Where the terms represents the concentrations of the acid and the ions.
By the ICE table, we can calculate the equilibrium concentrations,
CH3NH2 CH3NH3+
OH-
I 1.60 0
0
C -x +x
+x
--------------------------------------------------
E 1.60-x x
x
Kb = [OH-] [CH3NH3+] / [CH3NH2] = 3.4×10−4
Solving for x,
x = [OH-] = 0.023 M
pH = 14 + log 0.023 = 12.36
Therefore, the first option is the closest one.
Answer:
5.2
Explanation:
Calculate the pH of a 1.60 M CH₃NH₃Cl solution. Kb for methylamine, CH₃NH₂, is 3.7 × 10⁻⁴.
CH₃NH₃Cl is a strong electrolyte that ionizes according to the following equation.
CH₃NH₃Cl(aq) → CH₃NH₃⁺(aq) + Cl⁻(aq)
The concentration of CH₃NH₃⁺ will also be 1.60 M (Ca). CH₃NH₃⁺ is the conjugate acid of CH₃NH₂. We can find its acid dissociation constant (Ka) using the following expression.
Ka × Kb = Kw
Ka × 3.7 × 10⁻⁴ = 1.0 × 10⁻¹⁴
Ka = 2.7 × 10⁻¹¹
The acid dissociation of CH₃NH₃⁺ can be represented through the following equation.
CH₃NH₃⁺(aq) ⇄ CH₃NH₂(aq) + H⁺(aq)
For a weak acid, we can find the concentration of H⁺ using the following expression.
[H⁺] = √(Ka × Ca) = √(2.7 × 10⁻¹¹ × 1.60) = 6.6 × 10⁻⁶ M
The pH is:
pH = -log [H⁺] = -log (6.6 × 10⁻⁶) = 5.2
Calculate the ph of the resulting solution if 28.0 ml of 0.280 m hcl(aq) is added to
How many hydrogen atoms are in 5.20 mol of ammonium sulfide?
There are approximately 2.08 x 10^25 hydrogen atoms in 5.20 mol of ammonium sulfide.
Explanation:To determine the number of hydrogen atoms in 5.20 mol of ammonium sulfide (NH4)2S, we need to consider the molar ratio of hydrogen atoms to moles of ammonium sulfide. In the formula NH4)2S, there are 8 hydrogen atoms per 1 molecule. To find the number of hydrogen atoms, multiply the number of moles of ammonium sulfide by the Avogadro's number (6.022 x 1023 atoms/mol) and then multiply by the number of hydrogen atoms per molecule. So, the number of hydrogen atoms in 5.20 mol of ammonium sulfide is:
5.20 mol x 6.022 x 1023 atoms/mol x 8 atoms = 2.08 x 1025 hydrogen atoms.
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In [tex]5.20[/tex] moles of ammonium sulfide, there are approximately [tex]2.51 x 10^{25[/tex] hydrogen atoms.
To determine the number of hydrogen atoms in [tex]5.20[/tex] moles of ammonium sulfide [tex](\text{NH}_4)_2\text{S}[/tex], we must comprehend the compound's makeup.
First, note the formula of ammonium sulfide: [tex](\text{NH}_4)_2\text{S}[/tex].This indicates it contains two ammonium ions [tex]\text{(NH}_4^+)[/tex] and one sulfide ion ([tex]\text{S}^{2-}[/tex]).
Each ammonium ion ([tex]\text{NH}_4^+[/tex]) has 4 hydrogen atoms.Since there are two ammonium ions in each formula unit of ammonium sulfide, there are a total of 8 hydrogen atoms per formula unit of [tex](\text{NH}_4)_2\text{S}[/tex].
Now determine how many hydrogen atoms there are in total in [tex]5.20[/tex] moles of [tex](\text{NH}_4)_2\text{S}[/tex].Each mole of[tex](\text{NH}_4)_2\text{S}[/tex] contains Avogadro's number ([tex]6.022 \times 10^{23[/tex]) of formula units.
Thus, [tex]5.20[/tex] moles of [tex](\text{NH}_4)_2\text{S}[/tex] contain:
[tex](5.20 \, \text{moles}) \times (6.022 \times 10^{23} \, \text{formula units/mole}) = 3.13144 \times 10^{24} \, \text{formula units}[/tex]
Determine the pOH of a solution with the following 8.3*10^-7 M and classify the solution as acidic basic or neutral