Answer:
12.32 L.
Explanation:
The following data were obtained from the question:
Mass of CH4 = 8.80 g
Volume of CH4 =?
Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:
Mass of CH4 = 8.80 g
Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol
Mole of CH4 =?
Mole = mass/Molar mass
Mole of CH4 = 8.80 / 16
Mole of CH4 = 0.55 mole.
Finally, we shall determine the volume of the gas at stp as illustrated below:
1 mole of a gas occupies 22.4 L at stp.
Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.
Thus, 8.80 g of CH4 occupies 12.32 L at STP.
Answer:
0.1738 M
Explanation:
Let's consider the neutralization reaction between NaOH and HBr.
NaOH + HBr → NaBr + H₂O
The moles of HBr are:
14.76 × 10⁻³ L × 0.4122 mol/L = 6.084 × 10⁻³ mol
The molar ratio of NaOH to HBr is 1:1. In the endpoint, they have reacted completely, so the moles of NaOH were 6.084 × 10⁻³ mol before the reaction.
The molarity of NaOH is:
M = 6.084 × 10⁻³ mol / 35.00 × 10⁻³ L = 0.1738 M