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LUCKY_DIMON [66]
3 years ago
14

1. Baking soda and vinegar combined *

Chemistry
1 answer:
Alina [70]3 years ago
4 0

Answer:

dont know

Explanation:

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A piece of glass with a mass of 32.50 g specific heat of 0.840 J/g*°C and an initial temperature of 75 °C was dropped into a cal
elixir [45]
Hey there,

119.84 degrees c is your answer.

Hope I Helped!

Have a great day!


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The belief that you lose fat mass in a specific body part by concentrating exercise in that area is known as:
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Rate Law & Reaction Kinetics Chemistry 11 Chapter 17 1. For the reaction 3 ClO - (aq) → ClO - 3(aq) + 2 Cl - (aq) doubling t
ryzh [129]

Answer:

1) Rate = K [ClO⁻]²

2) Rate = K [C6H5N2Cl]

3) - Double the concentration of H3PO4.

- Double the concentration of I⁻

4) Check Explanation

Explanation:

A reaction's rate law is written as a product of the reaction's rate constant, k, and the concentration of the respective reactant(s) raised to the power of the order of reaction.

The order of a reaction with respect to a reactant is the power that the concentration of that specific reactant has in the rate law. It shows how dependent on each reactant , the rate of the reaction.

1) 3ClO⁻ (aq) → ClO³⁻ (aq) + 2 Cl⁻ (aq)

Doubling the concentration of ClO⁻ quadruples the initial rate of formation of ClO³⁻. What is the rate expression for the reaction?

Rate = k [ClO⁻]ⁿ

When [ClO⁻] is doubled, Rate is quadrupled, this shows that the reaction is second order with respect to the only reactant.

Rate = K [ClO⁻]²

2. The reaction

C6H5N2Cl (aq) + H2O (l) → C6H5OH (aq) + N2 (g) + HCl (aq) is first order in C6H5N2Cl and zero order in H2O. What is the rate expression?

Normally, the rate of reaction is equal to the rate constant multiplied by the each reactant's concentration raised ti the power of the order, so,

Rate = K [C6H5N2Cl]¹ [H2O]

Rate = K [C6H5N2Cl]

3. For the reaction

H3PO4 (aq) + 3I⁻ (aq) + 2H + (aq) → H3PO3(aq) + I³⁻(aq) + H2O

(l) the rate expression under certain conditions is R = k[H3PO4][I⁻][H⁺]² . What method(s) could be used if you want to double the reaction rate?

Rate = k[H3PO4][I⁻][H⁺]²

Indicating a first order relationship between the rate and the concentration of H3PO4 & I⁻ and second order with respect to H⁺.

So, any attempt to double the rate of reaction will entail a direct doubling of the one of the reactants with a first order relationship with the rate of reaction.

4. What is the overall order of reaction for each of the following.

a) R = k[NO2]² b) R = k c) R = k[H2][Br2] ½ d) Rate = k[NO]² [O2]

Note that overall order of a reaction is the sum of all the orders of the reactants that appear in the rate law.

a) R = k[NO2]²

Overall order is obviously 2.

b) R = k

Overall order is evidently 0.

c) R = k [H2] [Br2]^ ½

Overall order = 1 + ½ = (3/2)

d) Rate = k[NO]2 [O2]

Overall order = 2 + 1 = 3

Hope this Helps!!!

7 0
3 years ago
The following data was collected for the formation of ammonia (NH3) based on the following overall reaction: N2 + 3H2 = 2NH3 N2
notsponge [240]

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}

7 0
4 years ago
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