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Olegator [25]
3 years ago
10

How many moles of co2 will be produced from 72.0 g of ch4 assuming o2 is available in excess

Chemistry
1 answer:
valina [46]3 years ago
8 0

Moles of CO₂ produced will be 4.5

<u>Explanation:</u>

CH₄ + 2O₂ → CO₂ + 2H₂O

Molecular weight of CH₄, M = 16

given mass of CH₄, m = 72

moles of CH₄, n = given mass / molecular mass

n = \frac{m}{M} \\\\n = \frac{72}{16} \\\\n = 4.5

Therefore, number of moles of CH₄ present is 4.5

According to the balanced equation,

1 mole of CH₄ produces 1 mole of CO₂

4.5 moles of CH₄ will produce 4.5 moles of CO₂

Therefore, moles of CO₂ produced will be 4.5

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How many moles are there in 111.7 g of Fe?
Alenkinab [10]

Answer:

2

Explanation:

55.85 grams per mole.

times 2

111.7

6 0
2 years ago
PLZ HELP OR WILL FAIL!!
lianna [129]
Hey there!

Label A: Sublimation
Label B: Condensation
Label C: Melting

Remember sublimation is the transition of a substance directly from the solid to the gas state, without passing through the liquid state. Condensation is the conversion of a vapor or gas to a liquid. Melting is becoming liquefied by heat.

Hope this helps!
5 0
3 years ago
You have 5 cars and you double that amount.How much cars do you have.
Nostrana [21]

Answer:

10 cars

Explanation:

the double of 5 is 10

5+5=10

5 0
2 years ago
Part c how long does it take to raise the temperature of the air in a good-sized living room (3.00m×5.00m×8.00m) by 10.0∘c? note
postnew [5]
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4 0
3 years ago
An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the
san4es73 [151]

Answer:

the ionic radius of the anion r^- = 312.52 \ pm

Explanation:

From the diagram shown below :

The anion Cl^- is located at the corners

The cation Cs^+ is located at the body center

The Body diagonal length =  \sqrt{3 \ a }

∴ 2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a}  \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a

Given that :

\frac{r^+}{r^-} =0.84   (i.e the  ratio of the ionic radius of the cation to the ionic radius of

                 the anion )

0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a  \\ \\  1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a

Also ; a =  664 pm

Then :

r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm

Therefore,  the ionic radius of the anion r^- = 312.52 \ pm

4 0
3 years ago
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