2. What is the pressure in atm of 6.022 g CH, in a 30.0 L vessel at 129 C? A. 2.

Question

2. What is the pressure in atm of 6.022 g CH, in a 30.0 L vessel at 129 C? A. 2.42 B. 0.414 C. 12.4 D. 22.4 E. 6.62 . Select the one equation shown below for which the enthalpy change for the reaction is the standard heat of formation for the product. 3Mg(s) + N2(g) 2(g) Br2(1) D 02(g) + 2F2(g) A. Mg3N2(s) - B. C2H4(g) C2H6(g) + H2(g) -> 2HBr(g) + 02(g) 2F2(g)20F2C(g) > 20F2(g) N2(g) + 03(g) N203(g) . Which of the following is FALSE of real gases? A. Collisions are non-elastic. B. The volume is less than ideal because the gas particles themselves take up space. C. There are more deviations from ideal at high temperatures. D. The actual pressure is more than measured because there are attractions between gas particles

Explanation / Answer

2) we know that

moles = mass / molar mass

so

moles of CH4 = 6.022 / 16

moles of CH4 = 0.376375

now

PV = nRT

given

volume = 30

temperature = 129 C = 402 K

so

P x 30 = 0.376375 x 0.0821 x 402

P = 0.414 atm

so

the answer is B) 0.414

3)

we know that

for standard heat of formation

the product should be formed from individual elements in their standard state

also

one mole of product should be formed

B) C2H4 is not an element

C) one mole of product is not formed

D) one mole of product is not formed

E) oxygen should be in the form of 02 not 03

So

the correct answer is A)

3 Mg (s) + N2 (g) ---> Mg3N2 (s)

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