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muminat
4 years ago
6

6)

Chemistry
1 answer:
MrMuchimi4 years ago
7 0

Answer:

(4) Water, stirring, and filtering  

Explanation:

The added water will dissolve the sugar but not the sand. When you filter the mixture, the sand will be trapped in the filter paper and the dissolved sugar will pass through the pores of the paper.

(1) and (2) are wrong. You don't separate a mixture of sugar and sand by adding more sand. Furthermore, neither substance will boil at 100 °C.

(3) is wrong. You can dissolve the sugar in water but, if you boil the water away, the sugar and sand will still be together.

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What is the approximate mass of 25 cm3 of silver, in the density is 10.05g cm3?
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4 years ago
Student conducts an experiment to determine the enthalpy of solution for lithium chloride dissolved in water. the student combin
nirvana33 [79]

Answer:-  \Delta H_s_o_l_n=-37.2\frac{kJ}{mol}

Solution:- First of all we calculate the heat absorbed or released when the solute is added to the solvent. Here the solute is LiCl and the solvent is water.

To calculate the heat absorbed or released we use the formula:

q=ms\Delta T

q = heat absorbed or released

m = mass of solution

s = specific heat capacity

and \Delta T = change in temperature

mass of solution = mass of solute + mass of solvent

mass of solution = 5.00 g + 100.0 g = 105.0 g

(note:- density of pure water is 1 g per mL so the mass is same as its volume)

\Delta T = 33.0 - 23.0 = 10.0 degree C

s = 4.18\frac{J}{g.^0C}

Let's plug in the values in the formula and calculate q.

q = 105.0g(4.18\frac{J}{g.^0C})(10.0^0C)

q = 4389 J

To calculate the enthalpy of solution that is \Delta H_s_o_l_n we convert q to kJ and divide by the moles of solute.

moles of solute = 5.00g(\frac{1mol}{42.39g})

= 0.118 moles

q = 4389J(\frac{1kJ}{1000J}) = 4.389 kJ

\Delta H_s_o_l_n=\frac{4.389kJ}{0.118mol}

\Delta H_s_o_l_n = 37.2\frac{kJ}{mol}

Since the heat is released which is also clear from the rise in temperature of the solution, the sign of enthapy of solution will be negative.

So,  \Delta H_s_o_l_n=-37.2\frac{kJ}{mol}


4 0
3 years ago
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