According to law of definite proportion, for a compound, elements always combine in fixed ratio by mass.
The formula of compound remains the same, let it be a_{x}b_{y} where, a and b are two different elements.
Since, the ratio of mass remains the same , calculate the ratio of masses of element a and b in both cases
\frac{a}{b}=\frac{15}{35}=\frac{10}{y}
rearranging,
y=\frac{10\times 35}{15}=23.3
Thus, mass of b produced will be 23.3 g.
Answer:
= 9.28 g CO₂
Explanation:
First write a balanced equation:
CH₄ + 2O₂ -> 2H₂O + CO₂
Convert the information to moles
7.50g CH₄ = 0.46875 mol CH₄
13.5g O₂ = 0.421875 mol O₂
Theoretical molar ratio CH₄:O₂ -> 1:2
Actual ratio is 0.46875 : 0.421875 ≈ 1:1
If all CH₄ is used up, there would need to be more O₂
So O₂ is the limiting reactant and we use this in our equation
Use molar ratio to find moles of CO₂
0.421875 mol O₂ * 1 mol CO₂/2 mol O₂=0.2109375 mol CO₂
Then convert to grams
0.2109375 mol CO₂ = 9.28114 g CO₂
round to 3 sig figs
= 9.28 g CO₂
Oh my i really hope no one from my school sees my comments
Answer:
Explanation:
The question is asking, "At what temperature does the vapour pressure of water equal 3.4 atm?"
To answer this question, we can use the Clausius-Clapeyron equation:
Data:
p₁ = 1 atm; T₁ = 373.15C
p₂ = 3.4atm; T₂ = ?
R = 8.314 J·K⁻¹mol⁻¹
(The enthalpy of vaporization changes with temperature. Your value may differ from the one I chose.)
Calculation:
