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4 years ago

Iron iii chloride+ copper ii sulfate balanced equation

Chemistry

1 answer:

Bogdan [553]4 years ago

5 0

Answer:

Iron (iii) chloride reacts with copper (ii) sulfate to give iron sulfate and copper chloride.

Explanation:

The balanced equation of iron iii chloride+ copper ii sulfate is as follows.

$$2FeCl_{3}+3CuSO_{4} \to Fe_{2}(SO_{4})_{3}+3CuCl_{2}$

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3 years ago

Write the balanced complete ionic equation for the reaction hcho2(aq)+koh(aq)→

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3 years ago

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The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature: Time (s) (

harina [27]

Answer:

Part a: The rate of the equation for 1st order reaction is given as $Rate=k[H_2O_2]$

Part b: The integrated Rate Law is given as $[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c: The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d: Concentration after 4000 s is 0.043 M.

Explanation:

By plotting the relation between the natural log of concentration of $H_2O_2$ , the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.

Part a

Rate Law

The rate of the equation for 1st order reaction is given as

$Rate=k[H_2O_2]$

Part b

Integrated Rate Law

The integrated Rate Law is given as

$[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c

Value of the Rate Constant

Value of the rate constant is given by using the relation between 1st two observations i.e.

t1=0, M1=1.00

t2=120 s , M2=0.91

So k is calculated as

$-k(t_2-t_1)=ln{\frac{M_2}{M_1}}\\-k(120-0)=ln{\frac{0.91}{1.00}}\\k=\frac{-0.09431}{-120}\\k=7.8592 \times 10^{-4} s^{-1}$

The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d

Concentration after 4000 s is given as

$-k(t_2-t_1)=ln{\frac{M_2}{1.0}}\\-7.8592 \times 10^{-4}(4000-0)=ln{\frac{M_2}{1.00}}\\-3.1437=ln{\frac{M_2}{1.00}}\\M_2=e^{-3.1437}\\M_2=0.043 M$

Concentration after 4000 s is 0.043 M.

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3 years ago

HELPPP!!! What is the volume of a 1.31 moles sample of gas if the pressure is 904 mmHg and the temperature is 37. degrees Celsiu

Tju [1.3M]

Answer:

27.99 dm³

Explanation:

Applying

PV = nRT................ Equation 1

Where P = Pressure, V = Volume, n = number of mole, R = molar gas constant, T = Temperature.

From the question, we were aksed to find V.

Therefore we make V the subject of the equation

V = nRT/P................ Equation 2

Given: n = 1.31 moles, T = 37°C = 310K, P = 904 mmHg = (904×0.001316) = 1.1897 atm

Constant: R = 0.082 atm.dm³/K.mol

Substitute these values into equation 2

V = (1.31×310×0.082)/(1.1897)

V = 27.99 dm³

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3 years ago