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MatroZZZ [7]
3 years ago
12

Moist Brown solid Solid J K Oxygen Add solid L Heat Add dilute Solid M Colourless nygas N Black Precipitate P Bubble through CuS

O 4(aa) pitato 2017 H,SO (a) Identify J, K, L, M and N. (b) (i) Write an equation for the reaction that occurs between solid M and dilute sulphuric(VI) acid. (ii) Write an ionic equation for the reaction between N and copper(II) sulphate solution. (C) State and explain the observations made when N is bubbled through: (i) Iron(III) chloride solution. (ii) Dilute nitric(V) acid.​
Chemistry
1 answer:
Lerok [7]3 years ago
6 0

Answer:add solid l and k

Explanation:

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Which of the following reactions show a decrease in entropy? a. CaO(s) + H2O(l) Ca(OH)2 (s)
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The correct answer is <span>a. CaO(s) + H2O(l) Ca(OH)2 (s)
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7 0
3 years ago
Read 2 more answers
Consider the following reaction. SO2Cl2 → SO2 + Cl2. After collecting experimental data you found that plotting ln[SO2Cl2] vs. t
Nikitich [7]

Answer:

[SO_2Cl_2]_{600}= 0.0842 M

Explanation:

Some theoretical knowledge is required here. We should understand that whenever we plot the natural logarithm, ln, of a concentration vs. time and obtain a straight line, this indicates a first-order reaction. That said, since this is the case here, we have a first-order reaction with respect to SO_2Cl_2.

The linear equation has the following terms:

y = -0.000290t - 2.30

It is a linear form of the integrated first-order law equation:

ln[SO_2Cl_2]_t = -kt + ln[SO_2Cl_2]_o

Therefore, the rate constant, k, is:

k = 0.000290 s^{-1}

The natural logarithm of initial molarity is:

ln[SO_2Cl_2]_o = -2.30

Using the equation, we may substitute for t = 600 s and obtain the natural logarithm of the concentration at that time:

ln[SO_2Cl_2]_{600} = -0.000290 s^{-1}\cdot 600 s - 2.30 = -2.474

Take the antilog of both sides to find the actual molarity:

[SO_2Cl_2]_{600}=e^{-2.474} = 0.0842 M

4 0
3 years ago
The activation energy for a given reaction is 56 kj/mol. at what temperature would the rate constant be quadruple what is was at
Alex787 [66]

The rate equation is given as:

k = A e^(- Ea / RT)

 

Dividing state 1 and state 2:

k1/k2 = e^(- Ea / RT1) / e^(- Ea / RT2)

k1/k2 = e^[- Ea / RT1 - (- Ea / RT2)]

k1/k2 = e^[- Ea / RT1 + Ea / RT2)]

 

Taking the ln of both sides:

ln (k1/k2) = - Ea / RT1 + Ea / RT2

ln (k1/k2) = - Ea / R (1/T1 - 1/T2)

 

Since k2 = 4k1, therefore k1/k2 = ¼

 

ln (1/4) = [- (56,000 J/mol) / (8.314 J / mol K)] (1/273 K – 1/ T2)

2.058 x 10^-4 = 1/273 – 1/T2

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8 0
3 years ago
A catalyst will
scZoUnD [109]

Answer:

increase the chemical rate

7 0
3 years ago
What is plasma in chemistry​
faltersainse [42]

Explanation:

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hope this helps you

have a nice day

8 0
3 years ago
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