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Pachacha [2.7K]
3 years ago
11

HURRRYYYY

Chemistry
2 answers:
Sedaia [141]3 years ago
4 0

Answer:

1, 2, 1

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

Ludmilka [50]3 years ago
3 0

Answer:

1,2,1

Explanation:

just did the assigment

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Which is the isoelectronic of Na​
postnew [5]

Mg- is the isoelectronic of Na

6 0
3 years ago
If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?
Semenov [28]

Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

<em>Consider the following reaction: </em>

<em>SO2Cl2 -----> SO2(g) + Cl2(g) </em>

<em>Kc= 2.99 x 10^-7 at 227 degrees celcius </em>

<em>If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?</em>

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

       SOCl2 ---------> SO2 + Cl2     Kc = 2.99x10⁻⁷

i)       0.168                  0        0

c)         -x                    +x       +x

e)     0.168-x                x         x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

This means then, that the concentration of Cl2 in equilibrium would be:

<em>[Cl₂] = 2.24x10⁻⁴ M</em>

5 0
3 years ago
At a certain concentration of H2 and NH3, the initial rate of reaction is 0.120 M / s. What would the initial rate of the reacti
mel-nik [20]

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

rate=k[H_2]^2[NH_3]

At a certain concentration of H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]

As we are given that:

Initial rate = 0.120 M/s

Expression for rate law for first observation:

0.120=k[H_2]^2[NH_3] ....(1)

Expression for rate law for second observation:

R=k(\frac{[H_2]}{2})^2[NH_3] ....(2)

Dividing 2 by 1, we get:

\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}

\frac{R}{0.120}=\frac{1}{4}

R=0.03M/s

Therefore, the initial rate of the reaction will be, 0.03 M/s

5 0
3 years ago
When you apply 1000 joules of energy to 50 grams of water its temperature changes to 30 degrees . What was the initial temperatu
expeople1 [14]

Answer:

25.2°C

Explanation:

Given parameters:

Energy applied to the water  = 1000J

Mass of water  = 50g

Final temperature  = 30°C

Unknown:

Initial temperature  = ?

Solution:

To solve this problem, we use the expression below:

            H  = m c Ф

H is the energy absorbed

m is the mass

c is the specific heat capacity

Ф is the change in temperature

     1000  = 50 x 4.184 x (30  -  initial temperature )

     1000  = 209.2(30 - initial temperature)

      4.78  = 30 - initial temperature

      4.78  - 30  = - initial temperature

         Initial temperature  = 25.2°C

7 0
3 years ago
245 g water sample initially at at 32 oC absorbs 17 kcal of heat. What is the final temperature of water?
steposvetlana [31]

Answer:62.66°C or 235.66K

Explanation:Q=McpT, the energy was given in calories so you first convert to Joules by multiplying the value in calories by 4.184J.

17*4.184=71.128kJ.

71.128kJ=mcpT

71.128kJ=245*4.187*(T-Tm)

Tm is the final temperature of the mixture. The T is the temperature given which should be converted to Kelvin by adding 273...T=32+273=305K.

71128J=245*4.187*(305-Tm)

71128=312873.575-1025.815Tm

1025.815Tm=312873.575-71128

1025.815Tm=241745.58

Tm=241745.58/1025.815

Tm=235.66K

4 0
3 years ago
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