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oksano4ka [1.4K]
4 years ago
14

A birthday balloon had a volume of 14.1 L when the gas inside was at a temperature of 13.9 °C. Assuming no gas escapes, what is

its volume when the balloon warms up to 22.0 °C?
Chemistry
1 answer:
bixtya [17]4 years ago
8 0

Answer:

14.5L

Explanation:

The following data were obtained from the question:

V1 = 14.1L

T1 = 13.9°C = 13.9 + 273 = 286.9K

T2 = 22°C = 22 + 273 = 295K

V2 =?

Using charles' law: V1/T1 = V2 /T2, we can obtain the new volume as follows:

14.1/286.9 = V2 /295

Cross multiply to express in linear form

286.9 x V2 = 14.1 x 295

Divide both side by 286.9

V2 = (14.1 x 295) / 286.9

V2 = 14.5L

Therefore, the new volume = 14.5L

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3 years ago
How many moles of H2O2 are needed to react with 1.07 moles of N2H4?
I am Lyosha [343]

Answer:

2.14 moles of H₂O₂ are required

Explanation:

Given data:

Number of moles of H₂O₂ required = ?

Number of moles of N₂H₄ available = 1.07 mol

Solution:

Chemical equation:

N₂H₄  +   2H₂O₂       →   N₂ +  4H₂O

now we will compare the moles of H₂O₂ and N₂H₄

                          N₂H₄     :      H₂O₂  

                            1           :        2

                            1.07      :         2×1.07 = 2.14 mol

                   

6 0
3 years ago
what would you observe if you used a ballpoint pen instead of a pencil to mark the chromatography paper?
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It basically messes up the results

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If you use a ball point pen when doing a chromatogram, then the ink would separate as it is a mixture and run down the paper.

Graphite, or pencil lead however, is not an organic material and therefore will not be affected by common organic solvents used for thin-layer chromatography. Pen ink on the other hand will be readily absorbed by the solvent and will move up the plate.

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3 years ago
Identify the correct statement regarding the strength of chemical bonds.
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Calculate the volume occupied by 272g
cupoosta [38]

The answer for the following problem is mentioned below.

<u><em>Therefore volume occupied by methane gas is  184.78 × 10^-3 liters </em></u>

Explanation:

Given:

mass of methane(CH_{4}) = 272 grams

pressure (P) = 250 k Pa =250×10^3 Pa

temperature(t) = 54°C =54 + 273 = 327 K

Also given:

R = 8.31JK-1 mol-1 ,

Molar mass of  methane(CH_{4}) = 16.0​  grams

We know;

According to the ideal gas equation,

<u><em>P × V = n × R × T</em></u>

here,

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Therefore,

250×10^3 × V = 17 × 8.31 × 327

V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )

V = 184.78 × 10^-3 liters

<u><em>Therefore volume occupied by methane gas is  184.78 × 10^-3 liters </em></u>

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6 0
3 years ago
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