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galben [10]
3 years ago
9

A Gas OCCUPIES 525ML AT A PRESSURE OF 85.0 kPa WHAT WOULD THE VOLUME OF THE GAS BE AT THE PRESSURE OF 65.0 kPa

Chemistry
2 answers:
german3 years ago
8 0

Boyle's law of ideal gas: This law states that the volume of a gas is inversely proportional to its pressure at a constant temperature. Acc to this law we can write the relation of pressure and volume as:

PV=Constant

That means:

P_{1}V_{1}=P_{2}V_{2}

From that equation we can calculate Volume of gas at a certain pressure:

P₁=Initial pressure

V₁=Initial volume

P₂=Final pressure

V₂= Final volume

Here P₁, initial pressure is given as 85.0 kPa

V₁, initial volume is given as 525 mL

P₂, final pressure is 65.0 kPa

P_{1}V_{1}=P_{2}V_{2}

so,

V_{2}=85\times 525\div 65

=686 mL

Volume of gas will be 686 mL.

tatiyna3 years ago
4 0
The answer is 45.0 I think idk tho
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Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the fi
Savatey [412]

The question is incomplete, the complete question is;

Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the first heating, your crucible and contents weighs 17.51 g. After the second heating, your crucible and contents weighs 17.50 g.

What is the theoretical yield of sodium carbonate?

What is the experimental yield of sodium carbonate?

What is the percent yield for sodium carbonate?

Which errors could cause your percent yield to be falsely high, or even over 100%?

Answer:

See Explanation

Explanation:

We have to note that water is driven away after the second heating hence we are concerned with the weight of the pure dry product.

Hence;

From the reaction;

2 NaHCO3 → Na2CO3(s) + H2O(l) + CO2(g)

Number of moles of  sodium bicarbonate = 18.56 - 15.98 = 2.58 g/87 g/mol

= 0.0297 moles

2 moles of sodium bicarbonate yields 1 mole of sodium carbonate

0.0297 moles of 0.015 moles  sodium bicarbonate yields 0.0297 * 1/2 = 0.015 moles

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Experimental yield of sodium bicarbonate = 17.50 g - 15.98 g = 1.52 g

% yield = experimental yield/Theoretical yield * 100

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% yield = 96%

The percent yield may exceed 100% if the water and CO2 are not removed from the system by heating the solid product to a constant mass.

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Calculate the number of grams contained in each of the following number of moles. Report your answer with the appropriate number
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<u>Answers:</u>

a. 131.85 grams

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To calculate the number of grams in the number of moles given for each compound, consider the formula:

<em>Moles = No. of grams / molar mass</em>

Therefore, to find the gram it will become:

<em>No. of grams = moles x molar mass</em>

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No. of grams = moles x molar mass

No. of grams = 0.410 x 321.59 = 131.85 grams

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Molar mass of C2H4 = (12.01 x 2) + (1.01 x 4) = 28.06

No. of grams = moles x molar mass

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