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gizmo_the_mogwai [7]

4 years ago

7

A sample of gas is at 273°K and 2.00 atm with a volume of 6.00 liters. When the gas is compressed to 3.00 liters and the pressur

e increases to 3.00 atm, what is the new temperature of the sample?

1 answer:

horrorfan [7]4 years ago

3 0

Answer: 0.00488K

Explanation: using the general gas equation

(P1V1)/T1 =(P2V2)/T1

Substitute

2*6/273= 3*3/T

Simplify

T= 0.00488°K

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

adell [148]

Answer:

The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.

Explanation:

Step 1:

$CaC_2(s) + 2H_2O(g)\rightarrow C_2H_2(g) + Ca(OH)_2(s),\Delta H_1=414.0 kJ$ ...[1]

Step 2 :

$6C_2H_2(g) + 3CO_2(g) + 4H_2O(g)\rightarrow 5CH_2CHCO_2H(g) \Delta H_2=132.0kJ$ ..[2]

Adding 6 × [1] and [2]:

$6CaC_2(s) + 12H_2O(g)\rightarrow 6C_2H_2(g) + 6Ca(OH)_2(s)$

$6C_2H_2(g)+3CO_2(g)+16H_2O(g)\rightarrow 5CH_2CHCO_2H(g)$

we get :

$6CaC_2(s) + 8H_2O(g)+3CO_2(g)\rightarrow 5CH_2CHCO_2H(g)+ 6Ca(OH)_2(s),\Delta H'=?$

$\Delta H'=6\times \Delta H_1+\Delta H_2$

$\Delta H'=6\times 414.0 kJ+132.0kJ$

$\Delta H'=2,626 kJ$

Energy released on formation of 5 moles of acrylic acid = 2,626 kJ

Energy released on formation of 1 mole of acrylic acid:

$\frac{ 2,626 kJ}{5 } = 523.2 kJ$

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4 years ago

What are three things that living things need to live

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Food shelter oxygen

8 0

3 years ago

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3g of clean mg ribbon and 8g of clean copper metal was burnt separately in equal volume of air and both mass reacted completely

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3 years ago

A nugget of GOLD has a mass 9.66 gram and a volume of 0.5 cm3, It's density is<br>grams/cm3 *

dalvyx [7]

Answer:

P=19.32g/cm³

Explanation:

m=9.66g

v=0.5cm³

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3 years ago