Using PV = nRT, we can calculate the moles of the sample.
874 mmHg = 116,524 Pa
n = PV/RT
n = 116,524 x 294 x 10⁻⁶ / 8.314 x (140 + 273)
n = 9.98 x 10⁻³ mol
moles = mass / Mr
Mr = 0.271/9.98 x 10⁻³
Mr = 27.2
Mass of empirical formula = 14
Repeat units = 27.2 / 14 ≈ 2
Formula of substance:
C₂H₄
Combustion equation:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
1 mole produces 2 moles of CO₂, so 3 moles will produce 6 moles CO₂
<span>3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)
The unbalanced equation is:
ZnBr2 (aq) + Al (s) ==> AlBr3 (aq) +Zn (s)
First, count the atoms of each element on each side of the equation:
Zn 1,1
Br 2,3
Al 1,1
The zinc and aluminum, but the bromine doesn't match with 2 and 3. So look for the least common multiple of 2 and 3 which is 6 and adjust the quantities on both sides to have 6 bromine atoms on both sides. Do this by having 3 zinc bromide on the left and 2 aluminum bromide on the right, getting:
3 ZnBr2 (aq) + Al (s) ==> 2 AlBr3 (aq) +Zn (s)
Now check the atom counts again for both sides:
Zn 3,1
Br 6,6
Al 1,2
Now bromine matches, but zinc and aluminum doesn't. But it's easy enough to add an extra aluminum to the left and 2 more zinc to the right. Giving:
3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)
Now check the atom counts again:
Zn 3,3
Br 6,6
Al 2,2
And they match. So the balanced equation is:
3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)</span>
The molar mass of Na₂SO₄ -
2 x Na - 2 x23 = 46
1 x S - 1 x 32 = 32
4 x O - 4 x 16 = 64
total = 46 + 32 + 64 = 142 g/mol
the molarity of solution - 2.0 M
in 1 L of solution , 2.0 moles
Therefore in 2.5 L - 2 mol/L x 2.5 L = 5 mol
then the mass of Na₂SO₄ required = 142 g/mol x 5 mol = 710 g
<span>The law of proportion states that elements combine in whole number ratios. The gram readings for K are multiples of each other, both in grams and moles.
Let us compare the ratios:
</span>2.44 grams/1.22 grams = 2
<span>4.89 grams/2.44 grams = 2</span>
<span>Therefore, Potassium always combines with Oxygen in a ratio of 2 is to 1.</span>
First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.
PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂
The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>
