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mario62 [17]
4 years ago
9

Compressing a sample of hydrogen gas to one half its initial volume will have what effect on its pressure?

Chemistry
2 answers:
sammy [17]4 years ago
7 0

Answer:

The pressure will double

Explanation:

Hello,

In this case, by considering the Boyle's law which allows us to understand the pressure-volume behavior of a gas in an inversely proportional relationship:

P_1V_1=P_2V_2

It is said that the volume gets halved, thus:

V_2=\frac{1}{2}V_1

In such a way the pressure will result in:

P_2=\frac{P_1V_1}{V_2} =\frac{P_1V_1}{\frac{1}{2} V_1} \\\\P_2=2P_1

Therefore, the pressure will double.

Best regards.

avanturin [10]4 years ago
6 0
By Boyle's Law pV = constant, halfing the volume would require doubling the pressure to maintain the constant. Hence, pressure doubles.
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4 0
3 years ago
In another experiment, if 80 xo3 molecules react with 104 brz3 molecules how many br2 molecules will be produced which reactant
BaLLatris [955]

This is an incomplete question, here is a complete question.

The balanced chemical reaction is:

6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2

In another experiment, if 80 XO_3 molecules react with 104 BrZ_3 molecules. How many Br_2 molecules will be produced which reactant will be used up in the reaction.

Answer : The number of molecules of Br_2  will be, 52 molecules and BrZ_3 reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.

Explanation :

The balanced chemical reaction is:

6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2

First we have to determine the limiting reagent.

From the balanced reaction we conclude that,

As, 8 molecules of BrZ_3 react with 6 molecule of XO_3

So, 104 molecules of BrZ_3 react with \frac{104}{8}\times 6=78 molecule of XO_3

From this we conclude that, XO_3 is an excess reagent because the given moles are greater than the required moles and BrZ_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the molecules of Br_2

From the reaction, we conclude that

As, 8 molecules of BrZ_3 react to give 4 molecules of Br_2

So, 104 molecules of BrZ_3 react to give \frac{104}{8}\times 4=52 molecules of Br_2

Hence, the number of molecules of Br_2  will be, 52 molecules and BrZ_3 reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.

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Ammonia, NH3 is a common base with Kb of 1.8 X 10-5. For a solution of 0.150 M NH3:
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The concentrations : 0.15 M

pH=11.21

<h3>Further explanation</h3>

The ionization of ammonia in water :

NH₃+H₂O⇒NH₄OH

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The concentrations of all species present in the solution = 0.15 M

Kb=1.8 x 10⁻⁵

M=0.15

\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}

\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21

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Answer:

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Explanation:

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