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Alja [10]
3 years ago
10

Magnesium hydroxide is only very slightly soluble in water. The reaction by which it goes into solution is: Mg(OH) 2 (s) rightle

ftharpoons Mg^ 2+ (aq)+2OH^ minus (a q) What will happen if O H minus is added to the solution and why?
Chemistry
1 answer:
prohojiy [21]3 years ago
3 0

Answer:

there will a definite decrease in solute solution

Explanation:

acid reaction acting upon negative charge.

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We dissolve 2.45 g of sugar in 200.0 g water. what is the mass percent of sugar in the solution?
LiRa [457]

 The  mass percent  of  sugar  in the   solution is 1.225%


      <u><em>calculation</em></u>

mass percent =  [(mass  of solute /mass  of solution ) x  100]

mass   of  solute (  mass of a substance that  dissolve in another substance)

   = 2.45 g

mass  of  solution( mass of a  substance that dissolves  a solute) =200.0 g


Therefore mass  percent = 2.45 g/200.0 g  x 100 = 1.225% of sugar in solution

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When 18.9 kJ is transferred to a gas sample in a constant volume adiabatic container with a calorimeter constant of 2.22 Kj/K, t
poizon [28]

Answer:

(a) Cgas = 0.125 kJ/k

(b) cgas = 0.25kJ/kg.K

(c) cm(gas) = 0.021kJ/mol.K

Explanation:

18.9 kJ is equal to the sum of the heat absorbed by the gas and the heat absorbed by the calorimeter.

Qcal + Qgas = 18.9 kJ  [1]

We can calculate the heat absorbed using the following expression.

Q = C . ΔT

where,

C is the heat capacity

ΔT is the change in the temperature

<em>(a) What is the heat capacity of the sample?</em>

From [1],

Ccal . ΔT + Cgas . ΔT = 18.9 kJ

(2.22kJ/K) × 8.06 K + Cgas × 8.06 K = 18.9 kJ

Cgas = 0.125 kJ/k

<em>(b) If the sample has a mass of 0.5 kilograms, what is the specific heat capacity of the substance?</em>

We can calculate the specific heat capacity (c) using the following expression:

c=\frac{C}{m} =\frac{0.125kJ/K}{0.5kg} =0.25kJ/kg.K

<em>(c) If the sample is Krypton, what is the molar heat capacity at constant volume of Krypton? The molar mass of Krypton is 83.8 grams/mole.</em>

The molar heat capacity is:

\frac{0.25kJ}{kg.K} .\frac{1kg}{1000g} .\frac{83.8g}{mol} =0.021kJ/mol.K

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