Given an initial cyclopropane concentration of 0.00560 m, calculate the concentration of cyclopropane that remains after 1.50 hours.

We need an equation that would relate the concentration of the original solution to that of the desired solution. To solve this we use the equation expressed as follows, 

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

M1V1 = M2V2

0.266 M x V1 = 0.075 M x 150 mL

V1 = 42.29 mL

Therefore, we need about 42.29 mL of the 0.266 M of lithium nitrate solution to make 150.0 mL of the 0.075 M lithium nitrate solution.

[tex]\boxed{{\text{42}}{\text{.3 mL}}}[/tex] of a 0.266 M [tex]{\text{LiN}}{{\text{O}}_{\text{3}}}[/tex] solution is required to make 150 mL of a 0.075 M [tex]{\text{LiN}}{{\text{O}}_{\text{3}}}[/tex] solution.

Further Explanation:

The concentration is the proportion of substance in the mixture. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The molarity equation is given by the following expression:

[tex]{{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}} = {{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}[/tex]                      …… (1)

Here,

[tex]{{\text{M}}_{\text{1}}}[/tex] is the molarity of the initial [tex]{\text{LiN}}{{\text{O}}_{\text{3}}}[/tex] solution.

[tex]{{\text{V}}_{_{\text{1}}}}[/tex] is the volume of the initial [tex]{\text{LiN}}{{\text{O}}_{\text{3}}}[/tex] solution.

[tex]{{\text{M}}_{\text{2}}}[/tex] is the molarity of the new [tex]{\text{LiN}}{{\text{O}}_{\text{3}}}[/tex] solution.

[tex]{{\text{V}}_{_{\text{2}}}}[/tex] is the volume of the new [tex]{\text{LiN}}{{\text{O}}_{\text{3}}}[/tex] solution.

Rearrange equation (1) to calculate [tex]{{\text{V}}_{\text{1}}}[/tex].

[tex]{{\text{V}}_{\text{1}}}=\frac{{{{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}}}{{{{\text{M}}_{\text{1}}}}}[/tex]                    …… (2)

The value of [tex]{{\text{M}}_{\text{1}}}[/tex] is 0.266 M.

The value of [tex]{{\text{M}}_{\text{2}}}[/tex] is 0.075 M.

The value of [tex]{{\text{V}}_{_{\text{2}}}}[/tex] is 150 mL.

Substitute these values in equation (2).

[tex]\begin{aligned}{{\text{V}}_{\text{1}}}&=\frac{{\left({{\text{0}}{\text{.075 M}}} \right)\left( {{\text{150 mL}}} \right)}}{{{\text{0}}{\text{.266 M}}}}\\&=42.29{\text{ mL}}\\&\approx 42.{\text{3 mL}}\\\end{aligned}[/tex]

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: molarity, LiNO3, 42.3 mL, molarity equation, volume, M1, M2, V1, V2, 150 mL, 0.075 M, 0.266 M, concentration, concentration terms.

To solve this problem, we should recall that the change in enthalpy is calculated by subtracting the total enthalpy of the reactants from the total enthalpy of the products:

ΔH = Total H of products – Total H of reactants

You did not insert the table in this problem, therefore I will find other sources to find for the enthalpies of each compound.

ΔHf CO2 (g) = -393.5 kJ/mol

ΔHf CO (g) = -110.5 kJ/mol

ΔHf Fe2O3 (s) = -822.1 kJ/mol

ΔHf Fe(s) = 0.0 kJ/mol

Since the given enthalpies are still in kJ/mol, we have to multiply that with the number of moles in the formula. Therefore solving for ΔH:

ΔH = [3 mol ( − 393.5 kJ/mol) + 1 mol (0.0 kJ/mol)] − [3 mol ( − 110.5 kJ/mol) + 2 mol ( − 822.1 kJ/mol)]

ΔH = 795.2 kJ

Copper is a metallic atomic solid , the atoms are arranged in a regular pattern, with the valence electrons being free to move throughout the structure.Thus, the correct option is metallic atomic solid.

Copper is an example of a metal, and metals typically exhibit metallic bonding, where the valence electrons form a "sea" of delocalized electrons, creating strong bonds between the metal atoms. This allows for the high electrical and thermal conductivity that metals are known for.

Metallic solids are compounds that are entirely comprised of metal atoms that are held together by metallic bonds.Metallic bonding is a type of intramolecular force of attraction that occurs between a lattice of positive ions and a "sea" of delocalized electrons.

Thus, the correct option is metallic atomic solid.

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Answer:

Moles of H2O produced = 13.8

Explanation:

Given:

Moles of KMnO4 reacted = 3.45

To determine:

moles of H2O produced

Explanation:

Given reaction:

2KMnO4 + 16HCl →2KCl + 2MnCl2 + 8H2O + 5Cl2

Based on the reaction stoichiometry:

2 moles of KMnO4 produces 8 moles of H2O

Therefore, moles of H2O produced when 3.45 moles of KMnO4 react is:

[tex]= \frac{3.45\ moles\ KMnO4 * 8\ moles\ H2O}{2\ moles\ KMnO4} = 13.8[/tex]

To find the minimum volume of chlorine gas required to react with 7.85 g of aluminum, we convert the mass of aluminum to moles, find the necessary moles of chlorine gas using the balanced equation, and then apply the ideal gas law to find the volume.

The question is asking about the volume of chlorine gas required to completely react with a given amount of aluminum. We know from the balanced equation that 2 moles of aluminum (Al) react with 3 moles of chlorine gas (Cl) to form 2 moles of aluminum chloride (AlCl₃). First, we've to convert the mass of aluminum to moles by dividing the mass 7.85g by the molar mass of aluminum (26.98 g/mol), giving approximately 0.291 mol.

From the equation, we know the mole ratio of Al to Cl2 is 2 to 3. Therefore, 0.291 moles of Al will require 0.437 mol of Cl₂. Next, we apply the ideal gas law (PV=nRT) to find the volume. Here, P=225 mmHg (which is 0.296 atmospheres), R=0.0821(atm L)/(mol K), T=298 K and n=0.437 mol.

Finally, solving for V in PV=nRT gives us V = nRT/P, approximating 11.08 L as the minimum volume of chlorine gas required to react.

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Answer:

Exothermic chemical reaction

Endothermic chemical reaction

An Exothermic chemical reaction releases energy, whereas an Endothermic chemical reaction requires an input of energy.

An Exothermic chemical reaction involves the release of heat(thermal energy) in a system to the surroundings. The enthalpy(heat) change which is ΔH decreases in this type of reaction

An Endothermic chemical reaction involves the absorption or input of heat in the form of thermal energy by the system from the surroundings. The enthalpy(heat) change which is ΔH increases in this type of reaction.  

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Answer:

1- Carbon dioxide in our breath comes from the carbon in our food.

2- All plants need carbon dioxide to survive.

3- About .04 percent of the atmosphere's air is carbon dioxide and  4.4       percent of our breath is carbon dioxide we breathe out more carbon          dioxide than we breathe in.

4- Some effects might include combustion with other gasses and it could also potentially kill all life.

Explanation:

The Henderson-Hasselbalch equation fails to provide accurate pH readings for excessively diluted buffer solutions because it ignores the self-dissociation that occurs in water. The pH of the solution is 8.65.

The Henderson-Hasselbalch equation establishes a connection between an acid's pKa (acid dissociation constant) and pH in aqueous solutions. When the concentration of the acid and its conjugate base, or the base and the corresponding conjugate acid, are known, the pH of a buffer solution can be determined with the use of this equation.

The expression used to calculate pOH is:

pOH = pKb + log  [Conjugate acid]/ [Weak base]

pKa + pKb = 14

pKb = 14 - pKa

pKb = 14 - 9.25

pKb = 4.75

pOH = 4.75 + log 0.12 / 0.03

pOH = 5.35

pH = 14 - pOH

pH = 14 - 5.35

pH = 8.65

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Answer:

Explanation:

The compounds in exercise 48 are:

a) P4O6,

b) Ca3 (PO4)2, and

c) Na2 H PO4

So, proceed with the calculus for each compound.

a) Molecular formula: P4O6

Molar mass: 4 * 31 g/mol + 6* 16g/mol = 220 g/mol

Number of moles in 1.00 grams of compound = mass in grams / molar mass =

= 1.00 g / 220 g/mol = 0.004545 mol

0.004545 mol of P4O6 contains 4 * 0.004545 =  0.01818moles of atoms of P.

=> 0.01818 moles * 6.022 * 10^23 atoms / mol = 1.095 * 10^ 22 atoms of P.

Answer: 1.095 * 10^22 atoms of P.

b) Ca3 (PO4)2

molar mass = 3 * 40.1 g/mol + 2 * 31.0 g/mol + 8 * 16 g/mol = 310.3 g/mol

number of moles in 1.00 g of Ca3 (PO4)2 = 1.00 g / 310.3 g/mol = 0.00322 mol

0.00322 mol of compound * 2 mol P / mol of compound = 0.00644 mol P

0.00644 mol P * 6.022 * 10^23 atom / mol = 3.878 * 10 ^ 21 atoms P

Answer: 3.878 * 10^21 atoms P

c) Na2 H PO4

molar mass = 2 * 23.0 g/mol + 1 g/mol + 31.0 g/mol + 4 * 16g/mol = 142.0 g/mol

number of moles = 1.00 g / 142.0 g/mol = 0.0070 moles Na2HPO4

=> 0.0070 moles P

=> 0.0070 * 6.022 * 10^23 = 4.215 * 10^21 atoms of P

Answer: 4.215 * 10^21 atoms P

Final answer:

To determine the number of atoms of phosphorus in a compound, you need to use the molar mass and Avogadro's number. Convert the mass of the compound to moles and then multiply by Avogadro's number to get the number of atoms.

Explanation:

The number of atoms of phosphorus present in a compound can be determined using the molar mass and Avogadro's number. We need to convert the mass of the compound to moles using its molar mass, and then multiply by Avogadro's number to get the number of atoms.

For example, if we have 1.00g of phosphorus pentoxide (P2O5), we can calculate the number of atoms of phosphorus by:

The result will be the number of atoms of phosphorus in 1.00g of P2O5.

The equation that represents the process of photosynthesis is: 

6CO2+12H2O+light->C6H12O6+6O2+6H2O

Photosynthesis is the process in plants to make their food. This involves the use carbon dioxide to react with water and make sugar or glucose as the main product and oxygen as a by-product. Since we are not given the mass of CO2 in this problem, we assume that we have 1 g of CO2 available. We calculate as follows:

1 g CO2 ( 1 mol CO2 / 44.01 g CO2 ) ( 12 mol H2O / 6 mol CO2 ) ( 18.02 g / 1 mol ) = 0.82 g of H2O is needed

However, if the amount given of CO2 is not one gram, then you can simply change the starting value in the calculation and solve for the mass of water needed.

Final answer:

Photosynthesis involves converting carbon dioxide and water into glucose and oxygen with sunlight. The balanced chemical equation indicates a 6:1 mole ratio of water to glucose. The mass of water consumed would be six times the molecular weight of water times the moles of glucose formed.

Explanation:

Photosynthesis is a fundamental biological process in which green plants use sunlight to convert carbon dioxide (CO₂) and water (H₂O) into glucose (C₆H₁₂O₆) and oxygen (O₂). The balanced chemical equation for photosynthesis is:

6CO₂ + 6H₂O + sunlight → C₆H₁₂O₆ + 6O₂

The question asks about the mass of water consumed in the reaction with carbon dioxide to form glucose during photosynthesis. To determine this, you need to know the mole ratio between water and glucose from the balanced equation, which is 6:1, meaning six molecules of water are needed to produce one molecule of glucose. To find the specific mass of water used, you would multiply the molecular weight of water (18.01528 g/mol) by the number of moles of water consumed (which is typically six times the moles of glucose formed).

Nitrogen fertilizers are added to soil to provide readily available forms of nitrogen that plants can use. However, when these fertilizers enter aquatic ecosystems through runoff, they can cause eutrophication and harm aquatic fauna.

Nitrogen fertilizers are added to soil despite the dangers they pose to aquatic ecosystems because atmospheric nitrogen is not readily available for plants to use. Most plants require nitrogen in the form of nitrate or ammonium, which can be obtained from fertilizers. However, when these fertilizers enter aquatic ecosystems through runoff, they can cause eutrophication, leading to the overgrowth of microorganisms and the depletion of dissolved oxygen levels, which can be harmful to aquatic fauna.

Final answer:

To determine the number of moles of NaOH from 20 grams, calculate the substance's molar mass (39.997 g/mol) and divide the given mass by the molar mass, resulting in 0.5001 moles of NaOH.

Explanation:

The question asks about calculating the moles of sodium hydroxide (NaOH) from its mass in grams. To find moles, one has to use the molar mass of the substance.

First, calculate the molar mass of NaOH:
Na (1 x 22.990 g/mol) + O (1 x 15.999 g/mol) + H (1 x 1.008 g/mol) = 39.997 g/mol.

Next, use this molar mass to find the number of moles:
Given: 20.0 g NaOH,
Desired: moles NaOH
Since 1 mole of NaOH weighs 39.997 g, then:
moles of NaOH = mass (20 g) ÷ molar mass (39.997 g/mol) = 0.5001 moles.

The one I would expect to have the largest dispersion forces would be the largest and heaviest molecule. This is evidenced by the fact that that molecule is a liquid at room temperature while all the others are gases.

C3H8 = This is propane and a gas at room temperature

F2 = Also a gas at room temperature

BeCl2 = This is a solid and forms an extended lattice in the form of Be-Cl-Be bridges therefore dispersion forces are not important

Therefore the answer to this is C12H26 which is a wax and a liquid at room temperature.

Answer:

C12H26

Among the substances listed (C12H26, Be, Cl2, C3H8, F2), the largest dispersion forces are expected in C12H26 due to its larger molecular size and weight. Dispersion forces are temporary shifts in electron density causing attraction between molecules and are much stronger in larger and heavier molecules.  Smaller molecules like Cl₂ and F₂ have weaker dispersion forces.

The substance expected to have the largest dispersion forces from the ones mentioned (C12H26, Be, Cl2, C3H8, F2) is C12H26 due to its large size and molecular weight. Dispersion forces, also known as London dispersion forces, are temporary shifts in electron density in non-polar molecules that result in attraction between molecules. This is typically stronger in larger and heavier molecules. As C12H26 is a larger, heavier, and more complex molecule than the others listed, it has more electrons, hence more shifting of electron density and stronger resultant dispersion forces.

For other compounds like Cl₂ and F₂, they are gases at room temperature, meaning that their dispersion forces are weaker. This is because dispersion forces influence the boiling and melting points of substances. Larger dispersion forces lead to higher melting and boiling points, which is also why C12H26, a component of diesel and other heavy oils, is a liquid at room temperature.

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The total Hhydr is:

Hhydr = – 336 kJ/mol + – 532.7 kJ/mol

Hhydr = - 868.7 kJ/mol

Therefore using the formula Hsol = –Hlat + Hhydr we can get Hlat.

– 58 kJ/mol = – Hlat + - 868.7 kJ/mol

- Hlat = 810.7 kJ/mol

Hlat = - 810.7 kJ/mol

ANSWER: 

Answer:D

Explanation:

We can calculate for temperature by assuming the equation for ideal gas law:

P V = n R T

Where,

P = pressure = 1.80 atm

V = volume = 18.2 L

n = number of moles = 1.20 moles

R = gas constant = 0.08205746 L atm / mol K

Substituting to the given equation:

T = P V / n R

T = (1.8 atm * 18.2 L) / (1.2 moles * 0.08205746 L atm / mol K)

T = 332.70 K

We can convert K unit to ˚C unit by subtracting 273.15 to Kelvin, therefore

T = 59.55 ˚C

The concentration of Na ion in the final solution is 0.66 M.

To calculate the concentration of Na ions in the solution, the moles of [tex]\rm Na_2SO_4[/tex] in the solution have been calculated.

Moles = molarity [tex]\times[/tex] volume (L)

The molarity of [tex]\rm Na_2SO_4[/tex] solution given is = 0.4

Volume of [tex]\rm Na_2SO_4[/tex] solution = 100 ml = 0.1 L.

Moles of [tex]\rm Na_2SO_4[/tex] = 0.4 [tex]\times[/tex] 0.1

Moles of [tex]\rm Na_2SO_4[/tex] = 0.04

The dissociation of [tex]\rm Na_2SO_4[/tex] will be as follows:

[tex]\rm Na_2SO_2\;\rightarrow\;2\;Na^+\;+\;SO_4^2^-[/tex]

1 mole of [tex]\rm Na_2SO_4[/tex] = 2 moles of Na ions.

0.04 moles of [tex]\rm Na_2SO_4[/tex] = 0.04 [tex]\times[/tex] 2

0.04 moles of [tex]\rm Na_2SO_4[/tex] = 0.08 moles of Na ions.

The moles of Na ions in NaCl solution will be:

Given, molarity of NaCl solution = 0.6 M

Volume of NaCl solution = 200 ml = 0.2 L.

Moles of Na ions in NaCl solution = 0.6 [tex]\times[/tex] 0.2

Moles of Na ions in NaCl solution = 0.12.

The dissociation of NaCl will be as follows:

[tex]\rm NaCl\;\rightarrow\;Na^+\;+\;Cl^-[/tex]

1 mole of NaCl = 1 mole of Na ions

0.12 moles of NaCl gives = 0.12 moles of Na ions.

Thus the total Na ions = Na ion in [tex]\rm Na_2SO_4[/tex] + Na ions in NaCl

Total moles of Na ions = 0.08 + 0.12

Total moles of Na ions = 0.2 moles.

The total volume of solution will be :

0.1 L [tex]\rm Na_2SO_4[/tex] + 0.2 L NaCl

= 0.3 L.

Concentration of Na ion in the final solution = [tex]\rm \dfrac{moles}{volume\;(L)}[/tex]

The concentration of Na ion in the final solution =  [tex]\rm \dfrac{0.2}{0.3}[/tex] M

Concentration of Na ion in the final solution = 0.66 M.

The concentration of Na ion in the final solution is 0.66 M.

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The solution would be like this for this specific problem:

Given:

66.0 g of carbon monoxide

In the given reaction moles of carbon are needed to produce 66.0 g of carbon monoxide is 2.35 moles.

Moles of any substance will be calculated as:

n = W/M, where

W = given mass

M = molar mass

Given chemical reaction is:

2C + O₂ → 2CO

Moles of 66g of CO = 66g / 28g/mol = 2.35 mol

2 moles of CO = produced by 2 moles of carbon

2.35 moles of CO = produced by 2.35 moles of carbon

Hence, required moles of carbon are 2.35 moles.

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Answer: The balanced chemical equation is written below.

Explanation:

A balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side is equal to the total number of individual atoms on the product side. These equations follow law of conservation of mass.

The chemical equation for the reaction of iron (III) oxide with carbon follows:

[tex]2Fe_2O_3(s)+3C(s)\rightarrow 3CO_2(g)+4Fe(l)[/tex]

By Stoichiometry of the reaction:

2 moles of solid iron (III) oxide reacts with 3 moles of pure carbon to produce 3 moles of carbon dioxide gas and 4 moles of molten iron

Hence, the balanced chemical equation is written above.

Since the sample size is below 30, in this case we use the t statistic. The formula for t score is:

t = (x – u) / (σ / sqrt n)

where,

x = the level l = unknown

u = sample mean = 120 mg / dl

σ = standard deviation = 20 mg / dl

n = sample size or number of results = 5

Using the standard distribution tables for t, we can find the value of t given the probability (P = 0.15) and degrees of freedom (DOF).

t  = 1.036

Going back to the formula for t score:

1.036 = (x – 120) / (20 / sqrt 5)

x = 129.27 mg / dl = l

Final answer:

To find the glucose level such that the probability of the mean of 5 tests being above it is 0.15, we calculate the reduced standard deviation for the mean of the tests, find the corresponding z-score for a cumulative probability of 0.85, and apply the formula for the mean's distribution. The level l is found to be 129.267 mg/dl.

Explanation:

The question asks for the level l such that there is only a 0.15 probability that the mean glucose level of 5 test results falls above l, given that Sheila's glucose level after a sugary drink follows a normal distribution with a mean (μ) of 120 mg/dl and a standard deviation (σ) of 20 mg/dl.

To solve for l, we first need to recognize that the distribution of the mean of several test results (in this case, 5) will also be normally distributed, with the same mean but with a reduced standard deviation that is equal to the original standard deviation divided by the square root of the number of tests (σ/√n). For 5 tests, the new standard deviation (σnew) is 20/√5 = 8.9443 mg/dl.

The next step involves determining the z-score that corresponds to a probability of 0.85 (since 1 - 0.15 = 0.85) in the standard normal distribution. Consulting a z-table or using statistical software, we find that the z-score corresponding to 0.85 is approximately 1.036. Using the formula l = μ + z σnew, we calculate:

l = 120 + (1.036)(8.9443) = 120 + 9.267 = 129.267 mg/dl.

Therefore, the level l such that there is a probability of only 0.15 that the mean glucose level of 5 test results falls above l is 129.267 mg/dl.