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Nonamiya [84]
3 years ago
7

What celsius temperature, t2, is required to change the volume of the gas sample in part a (t1 = 43 ∘c , v1= 1.13×103 l ) to a v

olume of 2.26×103 l ? assume no change in pressure or the amount of gas in the balloon?
Chemistry
1 answer:
mamaluj [8]3 years ago
6 0
In this case, you are given the temperature and volume of the gas. To answer this question, you will need to use PV=nRT formula. Since the number of molecule and pressure is not changed then you can put 1 in both equations. The formula will become:

PV=nRT 
1 V= 1 T
V=T
T/V= 1

Dont forget that the celcius unit should be converted into kelvin. Then the calculation to find the temperature would be: 
T1/V1 = T2/V2
(43+273)/ 1.13x10^3 = T2/ 2.26 x 10^3
T2= (2.26 * 10^3) / (1.13 * 10^3) * (43+273)
T2= 2* 316= 632° K= 359°C
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2 years ago
Number 39! Please explain step by step!!!
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Answer:

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Explanation:

Data

Weight = 2900 lb

Weight = ? N

mass = ? kg

1 kg ------- 10 N

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Use proportions to solve this problem

                        0.22 lb ------------------- 1 N

                        2900 lb -----------------  x

                         x = (2900 x 1) / 0.22

                         x = 2900 / 0.22

                        x = 13181.8 N

                          1 kg ----------------------- 10 N

                           x     ----------------------- 13181.8 N

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                           x = 1318.2 kg

                       

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