Answer:
1.) 13 g C₄H₁₀
2.) 41 g CO₂
Explanation:
To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.
Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)
Molar Mass (C₄H₁₀): 58.124 g/mol
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O
48 g O₂ 1 mole 2 moles C₄H₁₀ 58.124 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 13 g C₄H₁₀
48 g O₂ 1 mole 8 moles CO₂ 44.007 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 41 g CO₂
Answer:
Answer E.
For a collision to be completely elastic, there must be NO LOSS in kinetic energy.
We can go through each answer choice:
A. Since the ball rebounds at half the initial speed, there is a loss in kinetic energy. This is NOT an elastic collision.
B. A collision involving sticking is an example of a perfectly INELASTIC collision. This is NOT an elastic collision.
C. A reduced speed indicates that there is a loss of kinetic energy. This is NOT elastic.
D. The balls traveling at half the speed after the collision indicates a loss of kinetic energy, making this collision NOT elastic.
E. This collision indicates an exchange of velocities, characteristic of an elastic collision. We can prove this:
Let:
m = mass of each ball
v = velocity
We have the initial kinetic energy as:
KE = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2KE=21mv2+0=21mv2
And the final as:
KE = 0 + \frac{1}{2}mv^2 = \frac{1}{2}mv^2KE=0+21mv2=21mv2
Answer:
Adding sodium or potassium hydroxide in amounts sufficient to convert all the H2SO4 into Na2SO4 would approximately neutralize the solution. The error would be the result of the imbalance between the basicity of the hydroxide and the acidity of the bisulfate (HSO4) anion. An adjustment in concentration would have to be made to achieve an accurate approximate pH of 7. But then you didn’t ask how much we would need to add.
Explanation:
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