Answer:
The resulting molarity is 6M.
Explanation:
A dilution consists of the decrease of concentration of a substance in a solution (the higher the volume of the solvent, the lower the concentration).
We use the formula for dilutions:
C1 x V1 = C2 x V2
12 M x 0,5L = C2 x 1,0 L
C2= (12 M x 0,5 L)/1,0 L
<em>C2= 6 M</em>
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
The answer is 492.8 g
1. Calculate a number of moles of a sample.
2. Calculate a molar mass of C3H8.
3. Calculate a mass of the sample.
1. Avogadro's number is the number of units (atoms, molecules) in 1 mole of substance: 6.023 × 10²³ units per 1 mole
6.023 × 10²³ atoms : 1 mol =6.72 × 10²⁴ atoms : n
n = 6.72 × 10²⁴ atoms * 1 mol : 6.023 × 10²³ atoms = 1.12 × 10 mol = 11.2 mol
2. Molar mass (Mr) of C3H8 is sum of atomic masses (Ar) of its elements:
Ar(C) = 12 g/mol
Ar(H) = 1 g/mol
Mr(C3H8) = 3 * Ar(C) + 8 * Ar(H) = 3 * 12 + 8 * 1 = 36 + 8 = 44 g/mol
3. Mass (m) of a sample is number of moles (n) multiplied by molar mass (Mr) of C3H8:
m = n * Mr = 11.2 mol * 44 g/mol = 492.8 g
Answer:
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