Wile E. Coyote is holding a "HEAVY DUTY ACME™ ANVIL" on a cliff that is 40.0 meters high. The Roadrunner (beep-beep), who is 1.0
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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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Answer:
The value of tension on the cable T = 1065.6 N
Explanation:
Mass = 888 kg
Initial velocity ( u )= 0.8
Final velocity ( V ) = 0
Distance traveled before come to rest = 0.2667 m
Now use third law of motion =
- 2 a s
Put all the values in above formula we get,
⇒ 0 = - 2 × a ×0.2667
⇒ a = 1.2
This is the deceleration of the box.
Tension in the cable is given by T = F = m × a
Put all the values in above formula we get,
T = 888 × 1.2
T = 1065.6 N
This is the value of tension on the cable.
M <span>represent mass in kg
</span><span>v represents speed in m/s
</span><span>r represents radius in m
Now, just substitute these into the formula:
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</span><span>v represents speed in m/s
</span><span>r represents radius in m
Now, just substitute these into the formula:
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