Answer:
Rb+
hope it's right:)
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C3H8+ 5 O2 --> 3 CO2 + 4 H2O
44 g. --------> 72 g
33.3 g. --------> x

Answer: The theoretical yield of H2O is 54.5
Answer:
62.5 moles of O₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2C₈H₁₈ + 25O₂ —> 16CO₂ + 18H₂O
From the balanced equation above,
2 moles of C₈H₁₈ reacted with 25 moles of O₂.
Finally, we shall determine the number of mole of O₂ needed to react with 5 moles of C₈H₁₈. This can be obtained as shown below:
From the balanced equation above,
2 moles of C₈H₁₈ reacted with 25 moles of O₂.
Therefore, 5 moles of C₈H₁₈ will react with = (5 × 25) / 2 = 62.5 moles of O₂.
Thus, 62.5 moles of O₂ is needed for the reaction.
M=11.20 g
m(H₂)=0.6854 g
M(H₂)=2.016 g/mol
M(Mg)=24.305 g/mol
M(Zn)=65.39 g/mol
w-?
m(Mg)=wm
m(Zn)=(1-w)m
Zn + 2HCl = ZnCl₂ + H₂
m₁(H₂)=M(H₂)m(Zn)/M(Zn)=M(H₂)(1-w)m/M(Zn)
Mg + 2HCl = MgCl₂ + H₂
m₂(H₂)=M(H₂)m(Mg)/M(Mg)=M(H₂)wm/M(Mg)
m(H₂)=m₁(H₂)+m₂(H₂)
m(H₂)=M(H₂)(1-w)m/M(Zn)+M(H₂)wm/M(Mg)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
m(H₂)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
(1-w)/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
1/M(Zn)-w/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
w(1/M(Mg)-1/M(Zn))=m(H₂)/{M(H₂)m}-1/M(Zn)
w=[m(H₂)/{M(H₂)m}-1/M(Zn)]/(1/M(Mg)-1/M(Zn))
w=0.583 (58.3%)
Answer:

Explanation:
Since we are given the mass, specific heat, and change in temperature, we should use this formula for heat:

The substance's mass is 450.0 grams, the specific heat is 1.264 J/g°C, and the change in temperature is 7.1 °C.

Substitute the values into the formula.

Multiply the first 2 values together. The grams will cancel out.

Multiply again. This time, the degrees Celsius cancel out.

<u>4038.48 Joules</u> of heat energy are released.