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lora16 [44]
3 years ago
8

If 1.8 L of water is added to 2.5 L of a 7.0 M KOH solution, what is the molarity of the new solution?

Chemistry
1 answer:
e-lub [12.9K]3 years ago
6 0
Molarity= (number of moles of that substance) / (Volume of solution in litres)

7= (number of moles of HCl) / 0.05

Thus,

Number of moles of HCl = 7 x 0.05 = 0.35

The number of moles of HCl will remain unchanged.

Now, for new solution,

0.35= 0.35 / (New volume of solution in litres)

Thus,

New volume = 1 litre

Thus, amount of water added = 1- 0.05 litre = 5.0 M

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A sample of 14.5 g of sodium bicarbonate (NaHCO3) was dissolved in 100 ml of water in a coffee-cup calorimeter with no lid by th
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Answer:

The ΔH of the reaction is + 12.45 KJ/mol

Explanation:

Mass of water= 100ml = 100g. (You should always assume 1cm3 of water as 1g)

heat capacity of water = 4.18 Jk-1 Mol-1

Change in temperature = (19.86 - 25.00) = -5.14 K (This is an endothermic reaction because of the fall in temperature)

Molar mass of NaHCO3 = 84 g/mol

Mole of NaHCO3 = 14.5 / 84 = 0.173 mol

Step 1 : Calculate the heat energy (Q) lost by the water.

            Q = M x C x ΔT

            Q = -100 x 4.18 x (-5.14)

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            Q = 2.1485 K J

Step 2: Calculating the ΔH of the reaction?

          ΔH = Q / number of moles of NaHCO3

          ΔH = 2.1485 / 0.173

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3 years ago
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3 0
3 years ago
Find the percent ionization of a 0.337 m hf solution. the ka for hf is 3.5 x 10-4. 1.1 % 1.2 x 10-2 % 3.2 % 3.5 x 10-2 % 4.7 %
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To determine the percent ionization of the acid given, we make use of the acid equilibrium constant (Ka) given. It is the ration of the equilibrium concentrations of the dissociated ions and the acid. The dissociation reaction of the HF acid would be as follows:<span>

HF = H+ + F-

The acid equilibrum constant would be expressed as follows:

Ka = [H+][F-] / [HF] = 3.5 x 10-4

To determine the equilibrium concentrations we use the ICE table,
         HF             H+              F-
I      0.337           0                 0
C      -x              +x               +x
---------------------------------------------
E    0.337-x        x                   x 

3.5 x 10-4 = [H+][F-] / [HF] 
3.5 x 10-4 = [x][x] / [0.337-x] </span>

Solving for x,

x = 0.01069 = [H+] = [F-]

percent ionization = 0.01069 / 0.337 x 100  = 3.17%

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Answer:

c

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