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Margarita [4]
3 years ago
14

Demonstrate how you can prepare 250ml 0.25M hydrogen peroxide from a solution of 20g/100ml of hydrogen peroxide.​

Chemistry
1 answer:
Nookie1986 [14]3 years ago
6 0

Answer:

Determine the volume of the initial solution,  

V

1

, to produce the required solution. We do this by applying the dilution equation,

M

1

V

1

=

M

2

V

2

where M is the concentration and V is the volume, while the subscripts respectively depicts the initial and required solutions. Now, we must express the concentration of the initial solution in terms of moles per liter, wherein we use the conversion factors such as the molar mass, 34.015 g/mol, and 1000 mL/L. We proceed converting the concentration.

M

1

=

20

 

g

1000

 

m

L

÷

34.015

 

g

/

m

o

l

×

1000

 

m

L

/

L

≈

5.88

 

M

Now, we are given the following values:

M

2

=

0.25

 

M

V

2

=

250

 

m

L

We proceed with the solution by applying the dilution equation and then plugging in the given values.

M

1

V

1

=

M

2

V

2

V

1

=

M

2

V

2

M

1

V

1

=

(

0.25

 

M

)

(

250

 

m

L

)

5.88

 

M

V

1

≈

11

 

m

L

Therefore, we need to take  

11

 

m

L

of the initial solution and dilute it to the desired volume to produce the required solution.

that's the explanation but, this helps if ur in higher classes though :>

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Mass % of NaOH solution = 10 % (m/v)

Mass of NaOH required to prepare 1 L solution:

1 L solution * \frac{1000 mL}{1L} *\frac{10 g NaOH}{100 mL solution}

= 100 g NaOH

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When ionic compounds are added to water, they dissociate or break apart into ions. In the case of table salt, salt added to wate
Lostsunrise [7]

Answer:

b dissociate

Explanation:

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Which property controls the movement of groundwater by determining the rate at which water passes through the soil?
Dimas [21]

The answer is permeability

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3 0
4 years ago
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erik [133]

Answer:

1) The correct step in the scientific method that Victor did is Construct a hypothesis.

2) Given mass and density, volume is calculated as mass divided by density.

Explanation:

1) Before doing the assay and make a graph with the results obtained, Victor should think what he wants to prove, so he should make a hypoythesis to test with the assay.

2) The formula of density is

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4 0
3 years ago
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Methane, a component in natural gas, can be used as a fuel in combustion reactions. What is the value for ΔGnon (in kJ) for the
Oksanka [162]

<u>Answer:</u> The \Delta G for the reaction is -806.86 kJ

<u>Explanation:</u>

We are given:

\Delta H^o_{rxn}=-803kJ=-803000J      (Conversion factor:  1 kJ = 1000)

\Delta S^o_{rxn}=-4.05J/K

Temperature of the reaction = 293 K

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

\Delta G^o_{rxn}=\Delat H^o_{rxn}-T\Delta S^o_{rxn}

Putting values in above equation, we get:

\Delta G^o_{rxn}=-803000J-[(293K)\times (-4.05J/K)]=-801813.35J

For the given chemical equation:

CH_4(g)+2O_2(g)\rightleftharpoons CO_2(g)+2H_2O(g)

The expression for K_c is given as:

K_{c}=\frac{[H_2O]^2[CO_2]}{[CH_4][O_2]^2}

We are given:

[H_2O]=6.41M

[CO_2]=3.83M

[CH_4]=14.51M

[O_2]=9.27M

Putting values in above equation, we get:

K_c=\frac{(6.41)^2\times 3.83}{14.51\times (9.27)^2}

K_c=0.126

To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_c

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = -801813.35 J

R = Gas constant = 8.314J/K mol

T = Temperature = 293 K

K_c = equilibrium constant in terms of concentration = 0.126

Putting values in above equation, we get:

\Delta G=-801813.35J+(8.314J/K.mol\times 293K\times \ln(0.126))

\Delta G=-806859.46J=-806.86kJ

Hence, the \Delta G for the reaction is -806.86 kJ

8 0
3 years ago
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