CHEMISTRY COLLEGE

If the heat of combustion of carbon monoxide (CO) is −283.0 kJmole, how many grams of carbon monoxide must combust in order to release 2,500.kJ of energy?

Answers

Answer 1
Answer:

Answer:

247.4 g

Explanation:

Let's consider the thermochemical equation for the combustion of carbon monoxide.

CO(g) + 0.5 O₂(g) ⇒ CO₂(g)    ΔH°c = -283.0 kJ/mol

The moles of carbon monoxide required to release 2500 kJ (-2500 kJ) are:

-2500 kJ × (1 mol CO/-283.0 kJ) = 8.834 mol CO

The molar mass of CO is 28.01 g/mol. The mass corresponding to 8.834 moles of CO is:

8.834 mol × 28.01 g/mol = 247.4 g

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Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. A mixture of 41.0 g of magnesium and 175.0 g of iron(III) chloride is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.

Answers

Answer: The limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

Explanation:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • For magnesium:

Given mass of magnesium = 41.0 g

Molar mass of magnesium = 24 g/mol

Putting values in equation 1, we get:

\text{Moles of magnesium}=(41.0g)/(24g/mol)=1.708mol

  • For iron(III) chloride:

Given mass of iron(III) chloride = 175.0 g

Molar mass of iron(III) chloride = 162.2 g/mol

Putting values in equation 1, we get:

\text{Moles of iron(III) chloride}=(175g)/(162.2g/mol)=1.708mol

The chemical equation for the reaction of magnesium and iron(III) chloride follows:

3Mg+2FeCl_3\rightarrow 3MgCl_2+2Fe

By Stoichiometry of the reaction:

3 moles of magnesium reacts with 2 moles of iron(III) chloride

So, 1.708 moles of magnesium will react with = (2)/(3)* 1.708=1.114mol of iron(III) chloride

As, given amount of iron(III) chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, magnesium is considered as a limiting reagent because it limits the formation of product.

Moles of excess reactant left (iron(III) chloride) = [1.708 - 1.114] = 0.594 moles

Now, calculating the mass of iron(III) chloride from equation 1, we get:

Molar mass of iron(III) chloride = 162.2 g/mol

Moles of iron(III) chloride = 0.594 moles

Putting values in equation 1, we get:

0.594mol=\frac{\text{Mass of iron(III) chloride}}{162.2g/mol}\n\n\text{Mass of iron(III) chloride}=(0.594mol* 162.2g/mol)=96.35g

Hence, the limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

Grunge is a rock style from Detroit. True False

Answers

Answer:

FALSE

Explanation:

Grunge refers to the genre of rock music and the fashion inspired by it. It originated in the mid-1980s in Seattle, Washington State.

Grunge was described as the fusion of punk rock and heavy metal.

This genre of music became popular in the early mid-1990s and included lyrics based on the theme of emotional and social alienation, betrayal, abuse, trauma etc.    

The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K. A sample of C8H18 is placed in a closed, evacuated 537 mL container at a temperature of 339 K. It is found that all of the C8H18 is in the vapor phase and that the pressure is 68.0 mm Hg. If the volume of the container is reduced to 338 mL at constant temperature, which of the following statements are correct?a. No condensation will occur.
b. Some of the vapor initially present will condense.
c. The pressure in the container will be 100. mm Hg.
d. Only octane vapor will be present.
e. Liquid octane will be present.

Answers

Answer:

the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.

Hence,

b. Some of the vapor initially present will condense.

e. Liquid octane will be present.

Explanation:

Given that;

The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K

Initial volume of the container, V1 = 537 mL

Initial vapor pressure, P1 = 68.0 mmHg

Final volume of the container, V2 = 338 mL

Let us say that the final vapor pressure = P2  

From Boyle's law,

P2V2 = P1V1

P2 * 338 = 68.0  * 537

338P2 = 36516

P2 = 36516 / 338

P2 = 108.03 mmHg

 

Thus, the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.

Hence,

b. Some of the vapor initially present will condense.

e. Liquid octane will be present.

At 25.0 ⁰C the henry's law constant for hydrogen sulfide(H2S) gas in water is 0.087 M/atm. Caculate the mass in grams of H2S gas that can be dissolved in 400.0 ml of water at 25.00 C and a H2S partial pressure of 2.42atm.

Answers

Answer: The mass of hydrogen sulfide that can be dissolved is 2.86 grams.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the gas.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_(H_2S)=K_H* p_(liquid)

where,

K_H = Henry's constant = 0.087M/atm

p_(H_2S) = partial pressure of hydrogen sulfide gas = 2.42 atm

Putting values in above equation, we get:

C_(H_2S)=0.087M/atm* 2.42atm\n\nC_(H_2S)=0.2105M

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = 0.2105 M

Molar mass of hydrogen sulfide = 34 g/mol

Volume of solution = 400.0 mL

Putting values in above equation, we get:

0.2105M=\frac{\text{Mass of hydrogen sulfide}* 1000}{34g/mol* 400.0mL}\n\n\text{Mass of }H_2S=(0.2105* 34* 400)/(1000)=2.86g

Hence, the mass of hydrogen sulfide that can be dissolved is 2.86 grams.

How does electronegativity affect the polarity of the bond between twoatoms?
A. The more electronegative atom will form the positive pole of a
polar bond.
B. The more electronegative atom will form a nonpolar end of the
bond.
C. The more electronegative atom will make its end of the bond more
negative.
D. Electronegativity differences between the atoms will cancel out
bond polarity

Answers

The electronegativity affects the polarity of the bond between two atoms, as the more electronegative atom will make its end of the bond more negative. The correct option is C.

What is electronegativity?

Electronegativity is a charge that shows the ability of an element to gain electron pairs with other elements during bonding. Electronegativity is altered by the distance between the electron and the nuclei and the atomic number of the element.

Polarity is the state of the atomic body in which it has placed charges in an opposite way to the other atoms so that they can join together.

Thus, the correct option is C. The more electronegative atom will make its end of the bond more negative.

To learn more about electronegativity, refer to the link:

brainly.com/question/17762711

#SPJ5

Answer:

C. The more electronegative atom willl make its end of the

bond more negative

A P E X

Consider the reaction at 25 °C. H2O(l) ↔ H2O(g) ΔG° = 8.6 kJ/mol Calculate the pressure of water at 25 °C (Hint: Get K eq)

Answers

Answer:

\boxed{\text{23.4 mmHg}}

Explanation:

H₂O(ℓ) ⟶ H₂O(g)

K_{\text{p}} = p_{\text{H2O}}

\text{The relationship between $\Delta G^(\circ)$ and $K_{\text{ p}}$ is}\n\Delta G^(\circ) = -RT \ln K_{\text{p}}

Data:  

T = 25 °C

ΔG° = 8.6 kJ·mol⁻¹

Calculations:

T = (25 + 273.15) K = 298.15 K

\begin{array}{rcl}8600 & = & -8.314 * 298.15 \ln K \n8600 & = & -2478.8 \ln K\n-3.47 & = & \ln K\nK&=&e^(-3.47)\n& = & 0.0311\end{array}

Standard pressure is 1 bar.

p_{\text{H2O}} = \text{0.0311 bar} * \frac{\text{750.1 mmHg}}{\text{1 bar}} = \textbf{23.4 mmHg}\n\n\text{The vapour pressure of water at $25 ^(\circ)\text{C}$ is $\boxed{\textbf{23.4 mmHg}}$}
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