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makvit [3.9K]
3 years ago
11

4.81 (a) What volume of 0.115 M HClO4 solution is needed to

Chemistry
1 answer:
neonofarm [45]3 years ago
7 0
Hahaha ?bajajsisjbwisi sosiwisos jk its so long also please make sure to stan loona and bts also stream back door by stray kids
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Please help with this question!
Alex17521 [72]

Answer:

A) potential energy is stored energy. Kenetic energy is energy of motion.

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3 years ago
Convert 1 bromopropane to bromoethane.​
Dovator [93]

Answer:

The Dehydrohaogenation of 1-bromo propane with alcoholic KOH gives propene which on again hydrohalogenation with HBr gives 2-bromo propane due to Markonikove's rule for addition.

Explanation:

6 0
3 years ago
A reaction mixture initially contains 0.140 MCO and 0.140 MH2O. What will be the equilibrium concentration of CO?
Allushta [10]

Answer:

0.013 M

Explanation:

From the question, we can make the following deductions; we are given mixture that contains two compounds, that is A and B, 0.140 M CO and 0.140 M H2O respectively. Then, we are asked to find the equilibrium concentration of Carbonmonoxide,CO.

So, the equation for the reaction is given below;

CO + H2O <-----------------> CO2 + H2.

Initially: we have 0.14M of CO, 0.14M of H2O and zero (0) concentration of CO2 and H2.

At time,t = CO =0.14 - x , H2O = 0.14 - x, CO2 and H2 = x.

The above reaction consist of the forward reaction and the backward reaction.

Therefore, the equilibrium Concentration of CO;

(Since we are giving that Kc = 102). Then, Kc=  [CO2][H2] ÷ [CO][H2O]. Where Kc is the equilibrium constant.

Therefore, 102 = [x^2] / [0.14 - x]^2.

==> 10.1= x/0.14 - x.

====> 0.141 - 10.1 x = x.

x + 10.1 x = 0.141.

===> 11.1 x = 0.141.

===> x = 0.141 ÷ 11.1.

===> x = 0.127 M .

Then, at time,t CO = 0.14 - x.

= 0.14 - 0.127 = 0.013 M

8 0
3 years ago
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The addition of 250.0 J to 30.0 g of copper initially at 22.0°C will change its temperature to what final value? (Specific heat
WINSTONCH [101]

Answer:

Final temperature = 43.53^{\circ} C

Explanation:

Given that,

Heat added, Q = 250 J

Mass, m = 30 g

Initial temperature, T₁ = 22°C

The Specific heat of Cu= 0.387 J/g °C

We know that, heat added due to the change in temperature is given by :

Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\T_2=\dfrac{Q}{mc}+T_1

Put all the values,

T_2=\dfrac{250}{30\times 0.387}+22\\\\=43.53^{\circ} C

So, the final temperature is equal to 43.53^{\circ} C.

8 0
3 years ago
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