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kenny6666 [7]
3 years ago
15

A science teacher has a supply of 50% sugar solution and a supply of 80% sugar solution. How much of each solution should the te

acher mix together to get 105 ML of 60% sugar solution for an experiment
70 ML of the 50% solution and 35 ML of the 80% solution
35 ML of the 50% solution and 70 ML of the 80% solution
70 ML of the 50% solution and 70 ML of the 80% solution
35 ML of the 50% solution and 35 ML of the 80% solution
Chemistry
1 answer:
Monica [59]3 years ago
5 0
You can establish a system of two equation with two variables.

Varibles are:
V1 = volume of the 50% sugar solution
V2 = volumen of the 80% sugar solution

Equations:
Balance of sugar:

Sugar from 50% solution: 0.5*V1
Sugar from 80% solution: 0.8*V2
Sugar in the final solution (mix): 0.6 * 105 = 63

1) 0.5V1 + 0.8V2 = 63

Final volume = volume of 50% solution + volume of 80% solution

2) V1 + V2 = 105

From (2) V1 = 105 - V2

Substitue in (1)

0.5 (105 - V2) + 0.8 V2 = 63

52.5 - 0.5V2 + 0.8V2 = 63

0.3 V2 = 63 - 52.5

0.3 V2 = 10.5

V2 = 10.5/0.3
V2 = 35mL

V1 = 105 - 35 = 70 mL

Answer: 70 mL of the 50% solution and 35 mL of the 80% solution.
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Answer : The enthalpy of reaction (\Delta H_{rxn}) is, 67.716 KJ/mole

Explanation :

First we have to calculate the moles of AgNO_3 and HCl.

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\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole

Now we have to calculate the moles of AgCl formed.

The balanced chemical reaction will be,

AgNO_3(aq)+HCl(aq)\rightarrow AgCl(s)+HNO_3(aq)

As, 1 mole of AgNO_3 react with 1 mole of HCl to give 1 mole of AgCl

So, 0.005 mole of AgNO_3 react with 0.005 mole of HCl to give 1 mole of AgCl

The moles of AgCl formed  = 0.005 mole

Total volume of the solution = 50.0 ml + 50.0 ml = 100.0 ml

Now we have to calculate the mass of solution.

Mass of the solution = Density of the solution × Volume of the solution

Mass of the solution = 1.00 g/ml × 100.0 ml = 100 g

Now we have to calculate the heat.

q=m\times C\Delta T=m\times C \times (T_2-T_1)

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q = heat

C = specific heat capacity = 4.18J/g^oC

m = mass = 100 g

T_2 = final temperature = 24.21^oC

T_1 = initial temperature = 23.40^oC

Now put all the given values in the above expression, we get:

q=100g\times (4.18J/g^oC)\times (24.21-23.40)^oC

q=338.58J

Now  we have to calculate the enthalpy of the reaction.

\Delta H_{rxn}=\frac{q}{n}

where,

\Delta H_{rxn} = enthalpy of reaction = ?

q = heat of reaction = 338.58 J

n = moles of reaction = 0.005 mole

Now put all the given values in above expression, we get:

\Delta H_{rxn}=\frac{338.58J}{0.005mole}=6771.6J/mole=67.716KJ/mole

Conversion used : (1 KJ = 1000 J)

Therefore, the enthalpy of reaction (\Delta H_{rxn}) is, 67.716 KJ/mole

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In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

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Equation for the second dissociation of sulfuric acid:

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The expression of second equilibrium constant equation follows:

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We know that:

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So, equilibrium concentration of sulfate ion = x = 0.00608 M

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