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3 years ago

A student increases the temperature of a 673 cm^3 balloon from 275 K to 460 K.

Chemistry

1 answer:

Kruka [31]3 years ago

8 0

Answer:

1125.75 cm³

Explanation:

We are given;

Initial volume; V1 = 673 cm³

Initial temperature; T1 = 275 K

Final temperature; T2 = 460 K

From Charles law, we can find the new volume from;

V1/T1 = V2/T2

V2 = (V1 × T2)/T1

V2 = (673 × 460)/275

V2 = 1125.75 cm³

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Answer:

When this equation is solved, the two values of the unknown are 0.0643 and -0.082

Explanation:

Given

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Required

The values of x

We have:

$1.77 * 10^{-2} = \frac{x^2}{0.298 - x}$

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Open bracket

$0.52746 - 1.77x= 100x^2$

Rewrite as:

$100x^2 + 1.77x - 0.52746 =0$

Using quadratic formula:

$x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

Where:

$a = 100; b = 1.77; c = -0.52746$

So, we have:

$x = \frac{-1.77 \± \sqrt{1.77^2 - 4*100*- 0.52746 }}{2*100}$

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Split

$x = \frac{-1.77 + 14.63}{200}\ or\ x = \frac{-1.77 - 14.63}{200}$

$x = \frac{12.86}{200}\ or\ x = \frac{-16.40}{200}$

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