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Answer:
D) The heavier ball will have a higher temperature because the change of temperature is inversely proportional to mass.
Explanation:
As stated in the problem, the amount of heat released by each ball is
where
m is the mass of the ball
Cp is the specific heat of iron (so, it is equal for both balls)
is the change in temperature of each ball
In this problem, we are said that the amount of heat released by the two balls, Q, is the same. Cp is also the same: this means that the product must be the same for the two balls. So, the mass and the change in temperature are inversely proportional: therefore, the heavier ball will have a smaller change in temperature. And since both balls starts from the same temperature, 100 C, this means that the heavier ball will reach a higher temperature than the lighter ball.
The solution of Sulfuric Acid (H2SO4) has the following mole fractions:
- mole fraction (H2SO4)= 0.034
- mole fraction (H2O)= 0.966
To solve this problem the formula and the procedure that we have to use is:
- n = m / MW
- = ∑ AWT
- mole fraction = moles of A component / total moles of solution
- ρ = m /v
Where:
- m = mass
- n = moles
- MW = molecular weight
- AWT = atomic weight
- ρ = density
- v = volume
Information about the problem:
- m solute (H2SO4) = 17.75 g
- v(solution) = 100 ml
- ρ (solution)= 1.094 g/ml
- AWT (H)= 1 g/mol
- AWT (S) = 32 g/mol
- AWT (O)= 16 g/mol
- mole fraction(H2SO4) = ?
- mole fraction(H2O) = ?
We calculate the moles of the H2SO4 and of the H2O from the Pm:
MW = ∑ AWT
MW (H2SO4)= AWT (H) * 2 + AWT (S) + AWT (O) * 4
MW (H2SO4)= (1 g/mol * 2) + (32,064 g/mol) + (16 g/mol * 4)
MW (H2SO4)= 2 g/mol + 32 g/mol + 64 g/mol
MW (H2SO4)= 98 g/mol
MW (H2O)= AWT (H) * 2 + AWT (O)
MW (H2O)= (1 g/mol * 2) + (16 g/mol)
MW (H2O)= 2 g/mol + 16 g/mol
MW (H2O)= 18 g/mol
Having the Pm we calculate the moles of H2SO4:
n = m / MW
n(H2SO4) = m(H2SO4) / MW (H2SO4)
n(H2SO4) = 17.75 g / 98 g/mol
n(H2SO4) = 0.1811 mol
With the density and the volume of the solution we get the mass:
ρ(solution)= m(solution) /v(solution)
m(solution) = v(solution) * ρ(solution)
m(solution) = 100 ml * 1.094 g/ml
m(solution) = 109.4 g
Having the mass of the solution we calculate the mass of the water in the solution:
m(H2O) = m(solution) - m solute (H2SO4)
m(H2O) = 109.4 g - 17.75 g
m(H2O) = 91.65 g
We calculate the moles of H2O:
n = m / MW
n(H2O) = m(H2O) / MW (H2O)
n(H2O) = 91.65 g / 18 g/mol
n(H2O) = 5.092 mol
We calculate the total moles of solution:
total moles of solution = n(H2SO4) + n(H2O)
total moles of solution = 0.1811 mol + 5.092 mol
total moles of solution = 5.2731 mol
With the moles of solution we can calculate the mole fraction of each component:
mole fraction (H2SO4)= moles of (H2SO4) / total moles of solution
mole fraction (H2SO4)= 0.1811 mol / 5.2731 mol
mole fraction (H2SO4)= 0.034
mole fraction (H2O)= moles of (H2O) / total moles of solution
mole fraction (H2O)= 5.092 mol / 5.2731 mol
mole fraction (H2O)= 0.966
<h3>What is a solution?</h3>In chemistry a solution is known as a homogeneous mixture of two or more components called:
- Solvent
- Solute
Learn more about chemical solution at: brainly.com/question/13182946 and brainly.com/question/25326161
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