a pro golfer hits 45.0-gram golf ball, giving its speed of 75.0 m/s. what is the momenntum has the golfer given to the ball

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Answer 1

Answer: Momentum = mass x velocity. 45g = 0.045kg. 0.045 x 75 = 3.375 kgm/s

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The reaction between NaOH(aq) and H2SO4(aq) was studied in a constant-pressure calorimeter: 100.0 mL portions of 1.00 M aqueous NaOH and H2SO4, each at 24.0C, were mixed. The volume of the mixture was 200.0 mL. The maximum temperature achieved was 30.6C. Neglect the heat capacity of the calorimeter and the thermometer, and assume that the solution of products has a density of 1.00 g/mL and a specific heat capacity of 4.18 J/(g·K). Calculate H, the heat (enthalpy) of reaction, in kJ per mole of acid. (Hint: You may not have the equivalent amounts of NaOH and H2SO4 in this experiment.)

Answers

Answer:

110,4 kJ / mole of acid

Explanation:

The reaction of NaOH with H₂SO₄ is:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Moles of NaOH and H₂SO₄ are:

0,1000L×1,00M = 0,100 moles of NaOH and 0,100 moles of H₂SO₄. As 2 moles of NaOH reacts with 1 mole of H₂SO₄, moles of acid that react are 0,100/2 = 0,0500moles of acid.

The produced heat is:

Q = C×m×ΔT

Where C is specific heat capacity (4,18J/gK)

m is mass: 200,0mL×(1,00g/mL) = 200g

ΔT is change in temperature: 30,6°C - 24,0°C = 6,6°C = 6,6K

Thus, Q is:

Q = 4,18J/gK×200g×6,6K

Q = 5518J = 5,52kJ

ΔH in kJ per mole of acid:

5,52kJ / 0,0500moles of acid = 110,4 kJ / mole of acid

I hope it helps!

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