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Angelina_Jolie [31]
3 years ago
6

What’s the answer to this? (2.8 x 10^4) (5.05 x 10^-6)

Chemistry
1 answer:
Verizon [17]3 years ago
8 0
The answer is

14.14 x 10^-2
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How much of a 60.0 gram sample of Carbon-14 DECAYED after 17,100 if 5
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The following data was collected for the formation of ammonia (NH3) based on the following overall reaction: N2 + 3H2 = 2NH3 N2
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Answer :  The unit for the rate constant in the rate law for the formation of ammonia is, M^{-2}min^{-1}

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

N_2+3H_2\rightarrow 2NH_3

Rate law expression for the reaction:

\text{Rate}=k[N_2]^a[H_2]^b

where,

a = order with respect to N_2

b = order with respect to H_2

Expression for rate law for first observation:

0.0021=k(0.10)^a(0.10)^b ....(1)

Expression for rate law for second observation:

0.0084=k(0.10)^a(0.20)^b ....(2)

Expression for rate law for third observation:

0.0672=k(0.20)^a(0.40)^b ....(3)

Dividing 2 by 1, we get:

\frac{0.0084}{0.0021}=\frac{k(0.10)^a(0.20)^b}{k(0.10)^a(0.10)^b}\\\\4=2^b\\b=2

Dividing 3 by 1 and also put value of b, we get:

\frac{0.0672}{0.0021}=\frac{k(0.20)^a(0.40)^2}{k(0.10)^a(0.10)^2}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[N_2]^a[H_2]^b

\text{Rate}=k[N_2]^1[H_2]^2

Now, calculating the value of 'k' by using any expression.

0.0021=k(0.10)^1(0.10)^2

k=2.1M^{-2}min^{-1}

The value of the rate constant 'k' for this reaction is 2.1M^{-2}min^{-1}

That means, the unit for the rate constant in the rate law for the formation of ammonia is, M^{-2}min^{-1}

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4 years ago
How many moles of water are produced when 5 moles of hydrogen gas react with 2 moles of oxygen gas?
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Well a molecule of water is equal to 1 Oxygen and 2 Hydrogen each, so it'd be 2 molecules of water

8 0
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You have an unknownthat contains only one cation (it is oneof the twelve from this experiment).Your solution is colorless and od
algol [13]

Answer:

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Explanation:

This is the explanation by chemical reactions

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Pb2+(aq)  +  2Cl-(aq)  --->  PbCl2 (s) ↓

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