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Sati [7]
3 years ago
13

One electron has a mass of 1 amu 2 amu 0 amu 3 amu

Chemistry
1 answer:
gladu [14]3 years ago
5 0
One electron has a mass of approximately 1 amu
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Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

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:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

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

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
Which planet formed near the sub where the solar system l’s temperatures were very high
Harrizon [31]

Answer:

Mars

Explanation:

Terrestrial or inner planets like Mars and Venus were formed near the Sun where the solar system's temperatures were very high.

4 0
3 years ago
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The reaction is first order in cyclopropane and has a measured rate constant of k=3.36×10−5 s−1 at 720 k. if the initial cyclopr
sergey [27]

Answer:

0.0277 M.

Explanation:

The integral rate law of a first order reaction:

<em>Kt = ln ([A₀]/[A]),</em>

where, k is the rate constant of the reaction <em>(k = 3.36 × 10⁻⁵ s⁻¹)</em>,

t is the time of the reaction <em>(t = 235.0 min = 14100 s)</em>,

[A₀] is the initial concentration of cyclopropane <em>([A₀] = 0.0445 M)</em>

<em>∵ Kt = ln ([A₀]/[A]),</em>

∴ (3.36 × 10⁻⁵ s⁻¹)(14100 s) = ln (0.0445 M)/[A]

Taking the exponential of both sides:

1.6 = (0.0445 M)/[A]

<em>∴ [A] = (0.0445 M)/1.6 = 0.0277 M.</em>

<em />

6 0
3 years ago
1. Given below are the figures showing particles of a matter in its three
stepan [7]

Answer:the answer is c

Explanation:

6 0
3 years ago
For any point on earths surface its height above sea level is called its
DedPeter [7]
Altitude, If this helped please 5 star me 
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3 years ago
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