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What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

4 0

2 years ago

A calorimeter contains 500 g of water at 25°C. You place a hand warmer containing 100 g of liquid sodium acetate (NaAC) inside t

antoniya [11.8K]

The heat absorbed by the water is
Q = 500 (4.18) (32.2 - 25)
Q = 15048 J

The enthalpy of fusion of the sodium acetate is:
ΔHf = Q / m
ΔHf = 15048 / 100
ΔHf = 150.48 J/g

3 0

2 years ago

Read 2 more answers

During laparoscopic surgery , carbon dioxide gas is used to expand the abdomen to help create a larger working space. If 4.80 L

Studentka2010 [4]

Answer:

5.37 L

Explanation:

To solve this problem we need to use the PV=nRT equation.

First we calculate the amount of CO₂, using the initial given conditions for P, V and T:

P = 785 mmHg ⇒ 785/760 = 1.03 atm

V = 4.80 L

T = 18 °C ⇒ 18 + 273.16 = 291.16 K

1.03 atm * 4.80 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 291.16 K

We solve for n:

n = 0.207 mol

Then we use that value of n for another PV=nRT equation, where T=37 °C (310.16K) and P = 745 mmHg (0.98 atm).

0.98 atm * V = 0.207 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 310.16 K

And we solve for V:

V = 5.37 L

7 0

2 years ago

Calculate the molarity of the solution formed when 0.72 moles of CaBr2 is dissolved in 1.50 L water.

xxTIMURxx [149]

Answer:

The molarity of the formed CaBr2 solution is 0.48 M

Explanation:

Step 1: Data given

Number of moles CaBr2 = 0.72 moles

Volume of water = 1.50 L

Step 2: Calculate the molarity of the solution

Molarity of CaBr2 solution = moles CaBr2 / volume water

Molarity of CaBr2 solution = 0.72 moles / 1.50 L

Molarity of CaBr2 solution = 0.48 mol / = 0.48 M

The molarity of the formed CaBr2 solution is 0.48 M

4 0

2 years ago

Question 1

exis [7]

Answer: False

Explanation: adding heat to matter results in thermal energy.

4 0

2 years ago