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ArbitrLikvidat [17]
2 years ago
12

2. How many moles of hydrogen atoms are there in 154 mL of 0.18 M H2SO4? Write your

Chemistry
1 answer:
katrin [286]2 years ago
6 0

2. 0.05544 moles of hydrogen atom are present in 154 mL 0.18 M solution of H2SO4.

3. 15.2506 heat in Joules is absorbed by 150.0 mL of pure water that is heated from 21.2°C to  45.5°C.

4. 0.75 M is the concentration of Na+ ions in 25.0 mL of 1.50 M NaOH is reacted with 25.0 mL of  1.50 M HCI

Explanation:

Number of moles of H2SO4 can be calculated by the given volume and molarity from the formula:

molarity = \frac{number of moles}{volume of teh solution}

number of moles = molarity × volume of the solution of H2SO4

    number of moles = 0.18 × 0.154 litres

                                   = 0.02772 moles of H2SO4.

Since 1 mole of H2SO4 contains 2 moles of hydrogen

so, 0.02772 moles of H2SO4 will have x moles

\frac{1}{2} = \frac{0.0272}{x}

2 × 0.02772 = x

0.05544 moles of hydrogen atom are present in 154 mL 0.18 M solution of H2SO4.

3. The heat absorbed is calculated from the formula:

ΔH = cp × m × ΔT   ( ΔT = change in temperature in Kelvin, m in kg, cp= specific heat of water)

putting the values in formula:

ΔH = 4.184 × 0.15 × ( 318.65-294.35)

     =  4.184 × 0.15× 24.3

        = 15.2506 Joules of heat is absorbed.

4. The concentration of Na+ ions

the balanced equation is

NaOH + HCl⇒ NaCl + H20

from one mole of NaCl 1 mole of NaCl i.e one 1 mole of Na+ ions is formed.

number of moles of NaOH is calculated by the formula:

Molarity = \frac{number of moles}{volume of the solution}

number of moles = 0.025L × 1.50M

                             = 0.0375 moles of NaOH

so 1 mole of NaOH produces 1 mole of Na= ions

hence, 0.0375 moles produces x moles of Na+ ions

\frac{1}{1} = \frac{0.0375}{x}

moles of Na+ is produced.  

concentration of NaOH in 50 ml solution because NaOH and HCl of 25 ml reacted.

applying the molarity formula from above

Molarity =  \frac{0.0375}{0.05}

              =  0.75 M

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This is an incomplete question, here is a complete question.

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