Answer:
Atomic mass of unknown gas is 4u
Explanation:
Average atomic mass of a gas is defined as the sum of molecular mass * percentage of each component in the mixture.
In the problem:
50.71u = 28×49.16% + 159.8×19.85% + 44×9.95% + X×21.04%
<em>Where X is atomic mass of unknown gas</em>
50.71u = 49.86 + X×21.04%
0.8469 = X×21.04%
<em>4u = X</em>
Atomic mass of unknown gas is 4u
Answer:
<h3>"Pearson correlation coefficient"</h3>
The correlation coefficient, denoted by r, tells us how closely data in a scatterplot fall along a straight line. The closer that the absolute value of r is to one, the better that the data are described by a linear equation. If r =1 or r = -1 then the data set is perfectly aligned.
<h3>Hope this is fine for you</h3>
Stoichiomety:
1 moles of C + 1 mol of O2 = 1 mol of CO2
multiply each # of moles times the atomic molar mass of the compund to find the relation is weights
Atomic or molar weights:
C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol
Stoichiometry:
12 g of C react with 32 g of O2 to produce 44 g of CO2
Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen
And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.
You cannot obtain 72 g of CO2 from 18 g of C.
May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.
Answer:
443 L of carbon dioxide
Explanation:
Combustion reaction: CH₄ (g) + 2O₂ (g) → CO₂(g) + 2H₂O(g)
We assume, that the oxygen is the excess reagent, so let's convert the mass of methane to moles. (mass (g) / molar mass)
Firstly we need to convert the mass from kg to g → 0.300 kg .1000 g / 1kg
300 g / 16 g/mol = 18.75 moles of methane.
Ratio is 1:1, so 18.75 moles of methane will produce 18.75 moles of CO₂
We apply the Ideal Gases Law to find the answer:
Firstly we need to convert the T°C to T°K → 15°C + 273 = 288 K
P . V = n . R. T → V = ( n . R . T ) / P We replace data:
V = (18.75 mol . 0.0821 L.atm/mol.K . 288K) / 1 atm → 442.8 ≅ 443 L
Answer:
this structure is n-pentane that's why ya yaha sa nai shuru howa
