Assuming conditions are at 25c and 1 bar 101.25 Kpa 1 mole of gas is equal to 22.4 L
So if you have 500L of ar gas the you have : 500L / 22.4 × 1 Mole = 22.3 moles of ar after accounting for sig figs..Hope that helps
Molecules that are similar enough to a neurotransmitter to bind to its receptor sites on a dendrite and block that neurotransmitter's effects are called <span>antagonists.</span>
Once the torch is lit, the acetylene flow must be increased until the flame stops smoking <span>before the oxygen is turned on for adjustment in order to keep the tip of the torch cool.
You should also note that while lighting the torch, you should keep the spark lighter near the tip but not covering it.</span>
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
