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2 years ago

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What volume of hydrogen iodide is produced when 118 liters of hydrogen gas react according to the following reaction? (All gases

are at the same temperature and pressure.) hydrogen(g) + iodine(s) hydrogen iodide(g) liters hydrogen iodide

1 answer:

Nadya [2.5K]2 years ago

4 0

Answer:

$V_{HI}=236LHI$

Explanation:

Hello,

In this case, for the given balanced chemical reaction:

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

Thus, since hydrogen and hydrogen iodide are in a 1:2 mole ratio, we can easily compute the yielded volume as shown below:

$V_{HI}=118LH_2*\frac{2molHI}{1molH_2} \\\\V_{HI}=236LHI$

Thus, is possible, due to the Avogadro's law which allows to relate moles and volume by a directly proportional relationship.

Regards.

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at what temperature in celsius would a gas have volume of 13.5 L at a pressure of 0.723 atm, if it had a volume of 17.8 L at a p

Alex

Answer:

initial temperature= $-3.31^{\circ}C$

Explanation:

Assuming that the given follows the ideal gas nature;

$P_1=0.723atm$

$V_1=13.5L$

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$P_2=0.612atm$

$V_2=17.8L$

$T_2=28 ^{\circ}C =273+28K=301K$

mole of gass will remain same at any emperature:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

putting all the value we get:

$T_2 =269.7K =-3.31 ^{\circ} C$

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Answer:

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What is one property of salts?

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Calculate the molarity of a kcl solution prepared by dissolving 0.898 moles of kcl in 250. ml of water

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2 years ago

Given the reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 505.9 kJ XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = + 199.3 kJ wh

eduard

Answer: -705.2 kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$X(s)+12O_2(g)\rightarrow XO(s)$ $\Delta H^0_1=-505.9kJ$ (1)

$XCO_3(s)\rightarrow XO(s)+CO_2(g)$ $\Delta H^0_2=+199.3kJ$ (2)

The final reaction is:

$X(s)+12O_2(g)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^0_3=?$ (3)

Reversing (2)

$XO(s)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^0_2'=-199.3kJ$ (2')

By adding (1) and (2')

$\Delta H^0_3=\Delta H^0_1+\Delta H^0_2'=-505.9kJ+(-199.3kJ)=-705.2kJ$

Hence $\Delta H^0=-705.2kJ$ .

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2 years ago