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My name is Ann [436]
2 years ago
8

Calculate the mass of water produced when 1.92 g of butane reacts with excess oxygen.

Chemistry
1 answer:
never [62]2 years ago
4 0
C4h10+6.5o2=4co2+5h2o
moles of butane=1.92/58=0.0331 moles
moles of water=0.1655 moles\
as the butane and water has 1 is to 5 molar ratio
0.1655=mass/18
mass=2.98 g
mass of water produced = 2.98 g
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Answer:

31.58 L

Explanation:

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

Initial mole (n₁) of gas = 1.90 moles

Initial volume (V₁) = 40 L

Final mole (n₂) = 1.90 – 0.40 = 1.5 moles

Final volume (V₂) =.?

The final volume of the gas can be obtained as follow;

V₁ / n₁ = V₂ / n₂

40 / 1.9 = V₂ / 1.5

Cross multiply

1.9 × V₂ = 40 × 1.5

1.9 × V₂ = 60

Divide both side by 1.9

V₂ = 60 / 1.9

V₂ = 31.58 L

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7 0
2 years ago
Fine the mole. 2.41 x 10^24 molecules CO2
vlada-n [284]

Answer:

<h2>4 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

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N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{2.41 \times  {10}^{24} }{6.02 \times  {10}^{23} }  \\  = 4.00322...

We have the final answer as

<h3>4 moles</h3>

Hope this helps you

5 0
3 years ago
The glycerol resulting from triacylglycerol metabolism can be converted to a form that can enter glycolysis. Identify the compou
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Answer:

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Answer:

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Explanation:

Given data:

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Formula weight of H₂O = 18.02 amu

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