Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
y’all i have this same question someone please help
Answer: The statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.
Explanation:
A strong acid upon dissociation gives a weak conjugate base. This can also be said as stronger is the acid, weaker will be its conjugate base or vice-versa.
Hydrofluoric acid is a strong base as it dissociates completely when dissolved in water.
For example, 
The conjugate base is 
which is a weak base.
Acetic acid is a weak acid as it dissociates partially when dissolved in water. So, the conjugate base of acetic acid is a strong base.
Thus, we can conclude that the statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.
The question is incomplete.
You need two additional data:
1) the original volume
2) what solution you added to change the volume.
This is a molarity problem, so remember molarity definition and formula:
M = n / V in liters: number of moles per liter of solution
To give you the key to answer this kind of questions, supppose the original volumen was 1 ml and that you added only water (solvent).
The original solution was:
V= 1 ml
M = 0.2 M
Using the formula for molarity, M = n / V
n = M×V = 0.2 M × (1 / 10000)l = 0.0002 moles
For the final solution:
n = 0.0002 moles
M = 0.04
From M = n / V ⇒ V = n / M = 0.002 moles / 0.04 M = 0.05 l
Change to ml ⇒ 0.05 l × 1000 ml / l = 50 ml. This would be the answer for the hypothetical problem that I assumed for you.
I hope this gives you all the cues you need to answer similar problems about molarity.
