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umka21 [38]
3 years ago
7

A student increases the temperature of a 200^3 cm balloon from 60°C to 120°C. What will the new volume of the balloon be? (Be ca

reful with units.)
A) 100 cm^3
B) 236 cm^3
C) 200 cm^3
D) 400 cm^3

(B is not the answer because I just took the test and i got it wrong)
Chemistry
1 answer:
Darina [25.2K]3 years ago
7 0

Explanation:

Given parameters:

Initial volume V1 = 200cm³

Initial temperature = 60°C

Final temperature = 120°C

Unknown:

Final volume = ?

Solution:

According to Charles law, at constant pressure, the volume of a given mass of gas is directly proportional to temperature.

mathematically;

    \frac{V_{1} }{T_{1} }  = \frac{V_{2} }{T_{2} }

1 and 2 are the initial and final states

V is the volume and T is the temperature

convert the temperature to kelvin

T1 = 60 + 273 = 333K

 T2 = 120 + 273 = 393K

Input the parameters;

\frac{200}{333}  = \frac{V_{2} }{393}

 final volume is 236cm³

learn more:

Boyle's law brainly.com/question/8928288

#learnwithBrainly

 

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13.9g of CuSO4•5H2O

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

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From the balanced equation above,

249.5g of CuSO4•5H2O produced 90g of H2O.

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From the balanced equation above,

249.5g of CuSO4•5H2O produced 90g of H2O.

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Therefore, 13.9g of CuSO4•5H2O is needed to produce 5g of H2O.

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