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castortr0y [4]
3 years ago
14

How many molecules are in 30 liters of methane (CH4) at STP?

Chemistry
1 answer:
mylen [45]3 years ago
4 0

Hello!

How many molecules are in 30 liters of methane (CH4) at STP ?

We have the following information:

Knowing that by Avogadro's Law for each mole of substance we have 6.02 * 10²³ molecules, it is known that in STP (Standard Temperature and Pressure) one mole of any gas equals 22.4 L, then:

22.4 L ----------------- 6.02*10²³ molecules

30 L -------------------- y molecules

\dfrac{22.4}{30} = \dfrac{6.02*10^{23}}{y}

multiply the means by the extremes

22.4*y = 30*6.02*10^{23}

22.4*y = 1.806*10^{25}

y = \dfrac{1.806*10^{25}}{22.4}

\boxed{\boxed{y = 8.0625*10^{23}\:molecules}}\:\:\:\:\:\:\bf\green{\checkmark}

Answer:

8.0625*10²³ molecules of methane

________________________  

\bf\green{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

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The question mentions a change in temperature from 25 to 50 °C. With that, the aim of the question is to determine the change in volume based on that change in temperature. Therefore this question is based on Gay- Lussac's Gas Law which notes that an increase in temperature, causes an increase in pressure since the two are directly proportional (once volume remains constant). Thus Gay-Lussac's Equation can be used to solve for the answer.
Boyle's Equation:     \frac{P_{1} }{T_{1} }    =   \frac{P_{2} }{T_{2} }
Since the initial temperature (T₁) is 25 C, the final temperature is 50 C (T₂) and the initial pressure (P₁) is 103 kPa, then we can substitute these into the equation to find the final pressure (P₂).
                      \frac{P_{1} }{T_{1} }    =   \frac{P_{2} }{T_{2} }∴ by substituting the known values,                 ⇒       (103 kPa) ÷ (25 °C)  =  (P₂) ÷ (50 °C)
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3 0
3 years ago
Pb(CH3COO)2 + H2S --> PbS + CH3COOH 1. How many moles are produced of PbS when 5.00 grams of Pb(CH3COO)2 is reacted with H2S?
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1. 0.0154mole of PbS

2. Double displacement reaction

Explanation:

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Pb(CH3COO)2 + H2S —> PbS + 2 CH3COOH

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From the equation,

1mole of Pb(CH3COO)2 produced 1mole of PbS.

Therefore, 0.0154mole of Pb(CH3COO)2 will also produce 0.0154mole of PbS

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