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natulia [17]
3 years ago
11

A sample of argon initially has a volume of 5.0 L and the pressure is 2 atm. If the final temperature is 30° C, the final volume

is 6 L, and the final pressure is 8atm, what was the initial temperature of the argon? Please give answer WITHOUT the unit of KELVIN and round to the tenth position (two places to the right of the decimal). Example: 25.14 not 25.4K

Chemistry
1 answer:
lbvjy [14]3 years ago
6 0
We have that every gas satisfies the fundamental gas equation, PV=nRT where P is the Pressure, V is the volume of the gas, n are the moles of the gas, R is a universal constant and T is the Temperature in Kelvin. We have that PV/T=nR and during our process, the moles of the gas do not change (no argon enters or escapes our sample). See attached.

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A gas occupies a volume of 650.0 mL when the pressure is 3.50 atm. What will the new volume be if the pressure is reduced to 1.6
Gemiola [76]

The new volume will be 1379 mL.

Explanation:

As per Boyle's law, the product of initial volume and initial pressure of any gas molecule is equal to the product of final volume and final pressure of those molecules.

So here the initial volume is 650 ml and the initial pressure is 3.50 atm. As the temperature is said to be constant, then this system will be obeying Boyle's law. So, the final pressure is given as 1.65 atm. As there is a reduction in the pressure, the volume of the gas is tend to get expanded.

P_{1} V_{1} = P_{2} V_{2}

So, (650*10^{-3}*3.50)=(1.65*V_{2})

V_{2} = \frac{650*10^{-3}*3.50}{1.65} = 1379 mL

So, the new volume of the gas on reduction in pressure is 1379 mL.

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There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In
KIM [24]

Answer:

ΔH = -470.4kJ

Explanation:

It is possible to sum 2 or more reactions to obtain the ΔH of the reaction you want to study (Hess's law). Using the reactions:

1. CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)ΔH = −414kJ

2. 6C2H2(g) + 3CO2(g) + 4H2O(g) → 5CH2CHCO2H(g)ΔH = 132kJ

6 times the reaction 1.

6CaC2(s) + 12H2O(l) → 6C2H2(g) + 6Ca(OH)2(s)ΔH = −414kJ*6 = -2484kJ

This reaction + 2:

6CaC2(s) + 3CO2(g) + 16H2O(l) →  + 6Ca(OH)2(s) + 5CH2CHCO2H(g) ΔH = -2484kJ + 132kJ = -2352kJ

As we want to calculate the net change enthalpy in the formation of just 1 mole of acrylic acid we need to divide this last reaction in 5:

6/5CaC2(s) + 3/5CO2(g) + 16/5H2O(l) →  + 6/5Ca(OH)2(s) + CH2CHCO2H(g) ΔH = -2352kJ / 5

<h3>ΔH = -470.4kJ</h3>

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3 years ago
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