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1 year ago

Predict the sign of ΔSsys for each process: (b) Hot air expands.

Chemistry

1 answer:

mars1129 [50]1 year ago

8 0

The sign of ΔSsys for hot air expands is positive.

Entropy (S) is the degree of randomness or disorderliness in a thermodynamic system. The randomness increases when the motion of molecules increases due to the addition of heat. It also increases when the distance between the molecules increases.

ΔS is a measurement of the change in entropy in a thermodynamic system. It is positive when entropy increases and negative when it decreases.

When the hot air expands the distance between the molecules of the gas increases and thus the randomness also increases. Entropy increases and the of ΔS becomes positive.

If you need to learn more about ΔSsys signs, click here

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Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of

bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

$O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol$ ..[1]

$O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol$ ..[2]

$NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol$ ..[3]

$NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?$ ..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

$2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}$

$2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol$

$2\times \Delta H^o_{4}=470.3 kJ/mol$

$\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol$

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

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

3 years ago

Iodine, I2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature

Stels [109]

<h3>Answer:</h3>

382.63 K

<h3>Explanation:</h3>

We are given;

Volume of Iodine as 71.4 mL
Mass of Iodine as 0.276 g
Pressure of Iodine as 0.478 atm

We are required to calculate the temperature of Iodine

We are going to use the ideal gas equation;

According to the ideal gas equation; PV = nRT, where R is the ideal gas constant, 0.082057 L.atm/mol.K.

Rearranging the formula;

T = PV ÷ nR

But, n, the number of moles = Mass ÷ Molar mass

Molar mass of iodine = 253.8089 g/mol

Thus, n = 0.276 g ÷ 253.8089 g/mol

= 0.001087 moles

Therefore;

T = (0.478 atm × 0.0714 L) ÷ (0.001087 moles × 0.082057)

= 382.63 K

Thus, the temperature of Iodine in Kelvin is 382.63 K

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Harlamova29_29 [7]

Answer:

Explanation:

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2 years ago

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