Answer:
The final volume will be "70.08 mL".
Explanation:
The given values are:
Molar mass,
M1 = 548 nM
or,
= 
M2 = 484 nM
or,
= 
Volume,
V1 = 61.9 mL
V1 = ?
By using the expression, we get
⇒ 
or,
⇒ 
By substituting the values, we get
Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol = 42. 6 L.
Answer: 43 L
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
Explanation:
We know that,
1 mile = 1609.34 m
We need to find how many meters are present in the 12.45 miles. To find it use unitary method as follows :
12.45 mile = 1609.34 × 12.45
12.45 mile=20036.283 meters
or
Hence, this is the required solution.
