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4 years ago

Which equation represents the combined gas law?

Chemistry

2 answers:

Pie4 years ago

8 0

Answer:

$\frac{P_1*V_1}{T_1}=\frac{P_2*V_2}{T_2}$

Explanation:

The combined gas law is the combination of three different laws applied to gases:

Gay-Lussac's Law:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Boyle-Mariotte's Law:

$P_1*V_1=P_2*V_2}$

Charles's Law:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

The result of the combination is:

$\frac{P_1*V_1}{T_1}=\frac{P_2*V_2}{T_2}$

mamaluj [8]4 years ago

4 0

((P1V1/T1)=(P2V2/T2))

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What is the name of the uppermost layer of the atmosphere? Why does it feel cold there, even though the temperature can be very

Aleonysh [2.5K]

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4 years ago

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<h3>Changes can occur to the physical properties of a substance, but the chemical composition of the substance remains the same.</h3>

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3 years ago

Describe what would be observed if a strip of magnesium ribbon was added to dilute hydrochloric acid.

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Answer:

A solid (s) magnesium strip or ribbon is added to hydrochloric acid, an aqueous solution (aq) . The reaction produces aqueous magnesium chloride and gaseous (g) hydrogen gas, which as you can see bubbles out of solution.

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3 years ago

Hydrazine, , emits a large quantity of energy when it reacts with oxygen, which has led to hydrazine used as a fuel for rockets:

Tcecarenko [31]

Answer:

$1.25~mol~H_2O$ and $0.627~mol~N_2$

Explanation:

Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ( $N_2H_4$ ) and oxygen ( $O_2$ ). So, we can start with the reaction between these compounds:

$N_2H_4~+~O_2~->~N_2~+~H_2O$

Now we can balance the reaction:

$N_2H_4~+~O_2~->~N_2~+~2H_2O$

In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for $N_2H_4$ and 31.99 g/mol for $O_2$ ):

$20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4$

$20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2$

In the balanced reaction we have 1 mol for each reagent (the numbers in front of $O_2$ and $N_2H_4$ are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is $N_2H_4$ .

With this in mind, we can calculate the number of moles for each product. In the case of $N_2$ we have a 1:1 molar ratio (1 mol of $N_2$ is produced by 1 mol of $N_2H_4$ ), so:

$0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2$

We can follow the same logic for the other compound. In the case of $H_2O$ we have a 1:2 molar ratio (2 mol of $H_2O$ is produced by 1 mol of $N_2H_4$ ), so:

$0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O$

I hope it helps!

4 0

4 years ago

Helium has the lowest boiling point of any substance, at 4.2 k. part a what is this temperature in âc?

Annette [7]

Answer is: temperature of helium is -296,95°C.
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T(He) = 4,2 K.
T(He) = 4,2 K - 273,15.
T(He) = -268,95°C.
The Celsius scale was based on 0°C for the freezing point of water and 100°C for the boiling point of water at 1 atm pressure.

6 0

3 years ago

Read 2 more answers