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Semenov [28]
3 years ago
10

As the temperature of a reaction increases , it is expected that the reacting particles collide

Chemistry
1 answer:
Oxana [17]3 years ago
6 0
It has been proven by Science that when the temperature of a reaction increases, the particles will gain energy and will collide faster and frequently.

Have a nice day! :)
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You are on an alien planet where the names for substances and the units of measures are very unfamiliar. Nonetheless, you obtain
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6.9 Sleps 1. (8)(13) = 15 2. 104/15 = 15/15 3. 6.93333333 4. 6.9 Sleps

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If 20.0 mL of a 0.0800 M HNO3, 35.0 mL of a 0.0800 M KSCN, and 40.0 mL of a 0.0800 M Fe(NO3)3 are combined, what is the initial
Tems11 [23]

Answer:

0.0295M

Explanation:

As you can see, in the mixture you have KSCN and other compounds. The KSCN in solution is dissolved in K⁺ ions and SCN⁻ ions. That means initial concentration of SCN⁻ ions is the same of KSCN, 0.0800M.

You are adding 35.0mL of this solution and the total volume of the mixture is 20.0mL + 35.0mL + 40.0mL = 95.0mL.

That means you are diluting your solution 95.0mL / 35.0mL = 2.714 times.

And the concentration of SCN⁻ is:

0.0800M / 2.714 =

<h3>0.0295M </h3>

4 0
3 years ago
HELP HELP HELP PLSSS
andrew11 [14]

Answer:

the answer is c kept in blue and with light

7 0
3 years ago
How many mol of phosphorus are obtained if 74.85 mol of carbon is used​
mixer [17]

Answer:

34.73

Explanation:

6 0
3 years ago
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
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