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Ganezh [65]
3 years ago
9

50cm3 of a mixture of methane and hydrogen were mixed with excess oxygen and exploded, the product after cooling to the original

room temperature were treated with excess koH solution when the volume decrease by 40cm3 , calculate the composition of the original mixture
Chemistry
1 answer:
marin [14]3 years ago
8 0

Answer:the initial composition of the reactants is

40cm^3 of CH4

40cm^3of H2

100cm^3 of H2O

Explanation:

Balanced reaction is

CH4 +H2+5/2O2______

CO2 +3H2O

Excess KOH at room temperature absorbs CO2 whose volume is given by 40cm^3 i.e the volume by which the solution decreases

So using Gay lussac combining ratio which states that gases combine in volumes that are in simple ratio to each other if gases.

Since CO2 in the equation is 1 mole

Means 1mole represent 40cm^3

So CH4:H2:O2 are in ratio of 1:1:5/2=(40:40:100)cm^3 respectively.

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If there are 1.55 x 1024 molecules of hydrogen peroxide (H2O2), what is the mass of the<br>sample? ​
bezimeni [28]

Answer:

87.54 g of H₂O₂

Explanation:

From the question given above, the following data were obtained:

Number of molecules = 1.55×10²⁴ molecules

Mass of H₂O₂ =.?

From Avogadro's hypothesis,

6.02×10²³ molecules = 1 mole of H₂O₂

Next, we shall determine the mass of 1 mole of H₂O₂. This can be obtained as follow:

1 mole of H₂O₂ = (2×1) + (2×16)

= 2 + 32

= 34 g

Thus,

6.02×10²³ molecules = 34 g of H₂O₂

Finally, we shall determine mass of H₂O₂ that contains 1.55×10²⁴ molecules. This can be obtained as follow:

6.02×10²³ molecules = 34 g of H₂O₂

Therefore,

1.55×10²⁴ molecules

= (1.55×10²⁴ × 34)/6.02×10²³

1.55×10²⁴ molecules = 87.54 g of H₂O₂

Thus, 87.54 g of H₂O₂ contains 1.55×10²⁴ molecules.

8 0
3 years ago
A laboratory analysis of an unknown sample yields 74.0% carbon, 7.4% hydrogen, 8.6% nitrogen, and 10.0% oxygen. What is the empi
kirza4 [7]
Let us assume that there is a 100g sample present. The respective mass of each element will then be:
C: 74 g
H: 7.4 g
N: 8.6 g
O: 10 g
Now, we divide each constituent's mass by its Mr to obtain the moles of each
C: (74 / 12) = 6.17
H: (7.4 / 1) = 7.4
N: (8.6 / 14) = 0.61
O: (10 / 16) = 0.625
Dividing by the smallest number:
C: 10
H: 12
N: 1
O: 1
Thus, the empirical formula is
C10H12NO
3 0
3 years ago
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