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lozanna [386]
3 years ago
6

HELPPP!!! What is the volume of a 1.31 moles sample of gas if the pressure is 904 mmHg and the temperature is 37. degrees Celsiu

s?
Chemistry
1 answer:
Tju [1.3M]3 years ago
4 0

Answer:

27.99 dm³

Explanation:

Applying

PV = nRT................ Equation 1

Where P = Pressure, V = Volume, n = number of mole, R = molar gas constant, T = Temperature.

From the question, we were aksed to find V.

Therefore we make V the subject of the equation

V = nRT/P................ Equation 2

Given: n = 1.31 moles, T = 37°C = 310K, P = 904 mmHg = (904×0.001316) = 1.1897 atm

Constant: R = 0.082 atm.dm³/K.mol

Substitute these values into equation 2

V = (1.31×310×0.082)/(1.1897)

V = 27.99 dm³

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saul85 [17]

<u>Answer:</u> The \Deltas H^o_{formation} for the reaction is -1406.8 kJ.

<u>Explanation:</u>

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.

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

The chemical reaction for the formation reaction of AlCl_3 is:

2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)    \Delta H^o_{formation}=?

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(2) H_2(g)+Cl_2(g)\rightarrow 2HCl(g)    \Delta H_2=-185kJ     ( ×  3)

(3) AlCl_3(aq.)\rightarrow AlCl_3(s)    \Delta H_3=+323kJ     ( ×  2)

(4) 2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)    \Delta H_4=-1049kJ

The expression for enthalpy of formation of AlCl_3 is,

\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]

Putting values in above equation, we get:

\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ

Hence, the \Deltas H^o_{formation} for the reaction is -1406.8 kJ.

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How many grams of CaF2 would be needed to produce 1.23 moles of F2?
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mass CaF2 = 78.07 g / mol * 1.23 mol

<span>mass CaF2 = 96.03 grams</span>

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