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FinnZ [79.3K]
3 years ago
14

Please use the values in the resources listed below instead of the textbook values. Under certain conditions the decomposition o

f ammonia on a metal surface gives the following data.[NH3] (M) 2.0 ✕ 10−3 4.0 ✕ 10−3 6.0 ✕ 10−3 Rate (mol/L/h) 1.5 ✕ 10−6 1.5 ✕ 10−6 1.5 ✕ 10−6 Determine the rate equation for this reaction. (Rate expressions take the general form: rate = k . [A]a . [B]b.)
Chemistry
1 answer:
Alex787 [66]3 years ago
4 0

Answer:

The rate equation for this reaction:

R=k[NH_3]^0

Explanation:

Decomposition of ammonia:

2NH_3\rightarrow N_2+3H_2

Rate law of the can be written as;

R=k[NH_3]^x

1) Rate of the reaction , when [NH_3]=2.0\times 10^{-3} M

1.5\times 10^{-6} M/s=k[2.0\times 10^{-3} M]^x..[1]

2) Rate of the reaction , when [NH_3]=4.0\times 10^{-3} M

1.5\times 10^{-6} M/s=k[4.0\times 10^{-3} M]^x..[2]

[1] ÷ [2]

\frac{1.5\times 10^{-6}M/s}{1.5\times 10^{-6}M/s}=\frac{k[2.0\times 10^{-3}M]^x}{k[4.0\times 10^{-3}M]^x}

On solving for x , we get ;

x = 0

The rate equation for this reaction:

R=k[NH_3]^0

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If HCl is a 0.05 M solution and 50 mL is used to titrate the NaOH, what is the molarity of NaOH if the flask contains 100 mL?
Liono4ka [1.6K]

Answer:

Molarity of NaOH = 0.025 M

Explanation:

Given data:

Molarity of  HCl = C₁ = 0.05 M

Volume of HCl = V₁= 50 mL

Molarity of NaOH = C₂=?

Volume of NaOH =V₂= 100 mL

Solution:

Formula:

C₁V₁  = C₂V₂

C₁ = Molarity of  HCl

V₁  = Volume of HCl

C₂ = Molarity of NaOH

V₂ = Volume of NaOH

Now we will put the values:

C₁V₁  = C₂V₂

0.05 M × 50 mL = C₂ × 100 mL

2.5 M.mL =C₂ × 100 mL

C₂  = 2.5 M.mL /100 mL

C₂  = 0.025 M

7 0
3 years ago
When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr
Amanda [17]

<u>Answer:</u> The value of K_c for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of N_2O_4 = 9.2 g

Molar mass of N_2O_4 = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M

For the given chemical equation:

                 N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>          0.20

<u>At eqllm:</u>     0.20-x        2x

We are given:

Equilibrium concentration of N_2O_4 = 0.057

Evaluating the value of 'x'

\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

[NO_2]_{eq}=2x=(2\times 0.143)=0.286M

[N_2O_4]_{eq}=0.057M

Putting values in above expression, we get:

K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435

Hence, the value of K_c for the given reaction is 1.435

6 0
3 years ago
The product of the nitration reaction will have a nitro group at which position with respect to the methyl group? Group of answe
zimovet [89]

Answer:

Mostly Para

Explanation:

First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).

Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.

Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.

Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.

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3 years ago
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Goryan [66]

Answer:

molecules

Explanation:

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7 0
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