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ddd [48]
3 years ago
11

Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) When 25.0 g of Zn reacts, how many L of H2 gas are formed at STP?

Chemistry
2 answers:
ludmilkaskok [199]3 years ago
7 0
Answer to this is 8.56L
irga5000 [103]3 years ago
4 0

Answer : The volume of hydrogen gas formed at STP are, 8.557 liters

Explanation : Given,

Mass of zinc = 25.0 g

Molar mass of zinc = 65.38 g/mole

First we have to calculate the moles of zinc.

\text{Moles of }Zn=\frac{\text{Mass of }Zn}{\text{Molar mass of }Zn}=\frac{25g}{65.38g/mole}=0.382moles

Now we have to calculate the moles of hydrogen gas.

The given balanced chemical reaction is,

Zn(s)+2HCl(aq)\rightarrow H_2(g)+ZnCl_2(aq)

From the balanced chemical reaction, we conclude that

As, 1 mole of Zn react to give 1 mole of hydrogen gas

So, 0.382 mole of Zn react to give 0.382 mole of hydrogen gas

Now we have to calculate the volume of hydrogen gas.

At STP,

As, 1 mole of hydrogen gas contains 22.4 L volume of hydrogen gas

So, 0.382 mole of hydrogen gas contains 0.382\times 22.4=8.557L volume of hydrogen gas.

Therefore, the volume of hydrogen gas formed at STP are, 8.557 liters

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Answer:

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7 0
3 years ago
Part E
AlekseyPX

Answer: similar, energy

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3 0
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What is the definition of lead in chemistry
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Answer:

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Explanation:

6 0
3 years ago
If the mass of a material is 104 grams and the volume of the material is 21 cm3, what would the density of the material be?
WITCHER [35]
Remembering that
d = m ÷ v

d = ?
m = 104 g
v = 21 cm³

Therefore:

d = 104 ÷ 21

<span>d = 4,95 g÷cm³</span>
6 0
3 years ago
What is the pressure inside a 750 mL can of deodorant that starts at 15 degrees Celsius and 1.0 atm if the temperature is raised
Sonbull [250]

The answer is: the pressure inside a can of deodorant is 1.28 atm.

Gay-Lussac's Law: the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.

p₁/T₁ = p₂/T₂.  

p₁ = 1.0 atm.; initial pressure

T₁ = 15°C = 288.15 K; initial temperature.

T₂ = 95°C = 368.15 K, final temperature

p₂ = ?; final presure.

1.0 atm/288.15 K = p₂/368.15 K.  

1.0 atm · 368.15 K = 288.15 K · p₂.  

p₂ = 368.15 atm·K ÷ 288.15 K.  

p₂ = 1.28 atm.  

As the temperature goes up, the pressure also goes up and vice-versa.  

6 0
3 years ago
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