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Marrrta [24]
3 years ago
15

A closed gas system initially has pressure and volume of 0.922 atm and 4.64L with the temperature unknown. If the same closed sy

stem has values of 1030 torr, 4.36 L, and 837K, what was the initial temperature in K?
Chemistry
1 answer:
JulsSmile [24]3 years ago
4 0

Answer:

606.09K

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the provided information in the question;

P1 = 0.922atm

P2 = 1030torr = 1.355atm

V1 = 4.64L

V2 = 4.36L

T1 = ?

T2 = 837K

Using P1V1/T1 = P2V2/T2

0.922 × 4.64/T1 = 1.355 × 4.36/837

4.278/T1 = 5.9078/837

Cross multiply

5.9078T1 = 837 × 4.278

5.9078T1 = 3580.69

T1 = 3580.69/5.9078

T1 = 606.09K

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Answer: 39.948 grams

Explanation:

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3 years ago
A chemistry student weighs out 0.027 kg of an unknown solid compound X and adds it to 550. mL of distilled water at 30.° C. Afte
QveST [7]

Answer:

1. Yes

2.The solubility of X is 34.55g/L

Explanation:

Solubility of  solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm^{3} or 1 Litre of water.

From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved

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if 19g is required to saturate 550mL at 30°C,

then\frac{19*1000}{550} will saturate 1L

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5 0
3 years ago
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree
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Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = T_1=33.9^oC=33.9+273K=306.9 K

Final temperature of the water = T_2

Specific heat capacity of water under these conditions =  c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)

-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC

The new temperature of the water bath 32.0°C.

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