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
Answer:
λ = 0.45×10⁻⁶ m
Explanation:
Given data:
Wavelength of blue light = ?
Frequency of blue light = 6.69×10¹⁴ s⁻¹
Solution:
Formula;
Speed of wave = wavelength × frequency
Speed of wave = 3.00×10⁸ m/s
by putting vales,
3.00×10⁸ m/s = λ × 6.69×10¹⁴ s⁻¹
λ = 3.00×10⁸ m/s / 6.69×10¹⁴ s⁻¹
λ = 0.45×10⁻⁶ m
Answer:the CO2 molecule has an excess of electron
The volume of hydrogen gas that evolved is calculated as follows
by use of ideal gas equation
that is PV = nRT
P=745 mm hg
V= ?
R(gas constant)= 62.36 L.mm hg/mol.k
T= 20 + 273 = 293 k
n=number of moles which is calculated as follows
find the moles of Na used
= 0.52/23=0.023 moles
write the reacting equation
2Na +2H2O =2NaOH +H2
by use of reacting ratio between Na : H2 which is 2:1 therefore the mole of H2 = 0.023/2 =0.0115 moles
by making the volume the subject of the formula
v=nRT/P
V= (0.0115 x 62.36 x 293) / 745 = 0.283 L
<h2>
Answer:</h2>
Average atomic mass of an element is the sum of the masses of its isotopes each multiplied by its natural abundance
