The _____melting point________ is the temperature at which a substance changes from solid to liquid; _______boiling point_________ is the temperature at which a substance changes from a liquid to as gas; _______vapourisation_________ is the process by which atoms of molecules leave a liquid and become a gas.
Answer:
B) 16 g
Explanation:
First we <u>convert 4 moles of O₂ into moles of H₂</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 4 mol O₂ *
= 8 mol H₂
Finally we <u>convert 8 moles of H₂ into grams</u>, using <em>its molar mass</em>:
- 8 mol H₂ * 2 g/mol = 16 g
Thus, the correct answer is option B).
Answer: The air will move more quickly around one side, making less pressure on that side of the ball
Explanation:
Answer:
1.332 g.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- At the same T and P and constant V (1.0 L), different gases have the same no. of moles (n):
<em>∴ (n) of CO₂ = (n) of C₂H₆</em>
<em></em>
∵ n = mass/molar mass
<em>∴ (mass/molar mass) of CO₂ = (mass/molar mass) of C₂H₆</em>
mass of CO₂ = 1.95 g, molar mass of CO₂ = 44.01 g/mol.
mass of C₂H₆ = ??? g, molar mass of C₂H₆ = 30.07 g/mol.
<em>∴ mass of C₂H₆ = [(mass/molar mass) of CO₂]*(molar mass) of C₂H₆</em> = [(1.95 g / 44.01 g/mol)] * (30.07 g/mol) =<em> 1.332 g.</em>
<em></em>
