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2 years ago

4. The decomposition of N20s in the gas phase is first order in N20s. If the concentration of N2Os drops from

Chemistry

1 answer:

arsen [322]2 years ago

4 0

Answer:

42.5 mins

Explanation:

Given that for a first order reaction;

ln[A] = ln[A]o - kt

[A] = 0.1 M

[A]o = 0.2 M

t= 14.2 mins

ln0.1 = ln0.2 - 14.2t

ln0.1 - ln0.2/-14.2=k

k= -2.303 -(-1.609)/-14.2

k= 0.0489 min-1

So;

ln0.025 =ln0.2 - 0.0489t

ln0.025 - ln0.2/-0.0489 =t

t= -3.689 - (-1.609)/-0.0489

t= 42.5 mins

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2 years ago

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3 years ago

c3H8 + 5O2 = 3CO2 + 4H2O When 44.0 grams of propane (C3H8) under goes complete combustion, how many grams of water will be produ

Umnica [9.8K]

<u>Answer:</u> 72 grams of water will be produced.

<u>Explanation:</u>

To calculate the number of moles, we use the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

Mass of propane = 44 grams

Molar mass of propane = 44 grams

Putting values in above equation, we get:

$\text{moles of propane}=\frac{44g}{44g/mol}=1mole$

For the reaction of combustion reaction of propane, the equation follows:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

By Stoichiometry of the reaction,

1 mole of propane produces 4 moles of water.

So, 1 mole of propane will produce = $\frac{1}{1}\times 4=4moles$ of water.

Now, to calculate the amount of water, we use equation 1, we get:

Molar mass of water = 18 g/mol

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6 0

3 years ago

What is the percent by mass of nahco3 in a solution containing 10 g of nahco3 dissolved in 400 ml of h2o?

Marat540 [252]

Answer is: the percent by mass of NaHCO₃ is 2,43%.
m(NaHCO₃) = 10 g.
V(H₂O) = 400 ml.
d(H₂O) = 1 g/ml.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 400 ml · 1 g/ml.
m(H₂O) = 400 g.
m(solution) = m(H₂O) + m(NaHCO₃).
m(solution) = 400 g + 10 g.
m(solution) = 410 g.
ω(NaHCO₃) = 10 g ÷ 410 g · 100%.
ω(NaHCO₃) = 2,43 %

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3 years ago

How many grams of carbon are present in 45.0 g of CCl4?

lisabon 2012 [21]

To determine the amount of a certain element in a compound, we use the ratio of the elements from the compound. We calculate is follows:

45.0 g CCl4 ( 1 mol CCl4 / 153.82 g CCl4 ) ( 1 mol C / 1 mol CCl4 ) ( 12.01 g C / 1 mol C ) = 3.5135 g carbon present

Hope this answers the question. Have a nice day.

5 0

3 years ago