Answer:
Explanation:
Given parameters:
Initial temperature T₁ = 25.2°C = 25.2 + 273 = 298.2K
Initial pressure = P₁ = 0.6atm
Final temperature = 72.4°C = 72.4 + 273 = 345.4K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we use an adaption of the combined gas law where the volume gas is fixed. This simplification results into:
where P and T are temperatures, 1 and 2 are initial and final temperatures.
Input the parameters and solve;
P₂ = 0.7atm
It would have to be paints consists of pigments,solvents, and binders. Once the [paint has been applied and has dried, the pigments are still able to determine the matched samples.
Answer:
Explanation:
<u>1) Balanced chemical equation:</u>
<u>2) Mole ratio:</u>
- 2 mol S : 3 mol O₂ : 2 mol SO₃
<u>3) Limiting reactant:</u>
n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂
n = 7.0 g / 32.065 g/mol = 0.2183 mol S
Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859
Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5
Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.
<u>4) Calcuate theoretical yield (using the limiting reactant):</u>
- 0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃
- x = 0.1875 × 2 / 3 mol SO₃ = 0.125 mol SO₃
<u>5) Yield in grams:</u>
- mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol = 10.0 g
<u>6) </u><em><u>Percent yield:</u></em>
- Percent yield, % = (actual yield / theoretical yield) × 100
- % = (7.9 g / 10.0 g) × 100 = 79%
Answer: a) 
b) 
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
a) Mass of Ba= 66.06 g
Mass of Cl = 34.0 g
Step 1 : convert given masses into moles.
Moles of Ba =
Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Ba = 
For O =
The ratio of Ba: Cl= 1:2
Hence the empirical formula is
b) Mass of Bi= 80.38 g
Mass of O= 18.46 g
Mass of H = 1.16 g
Step 1 : convert given masses into moles.
Moles of Bi =
Moles of O= 
Moles of H=
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Bi= 
For O =
For H=
The ratio of Bi: O: H= 1:3: 3
Hence the empirical formula is
Answer is: 0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K.
Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - <span>Henry's law constant.
</span>c - solubility of a gas at a fixed temperature in a particular solvent.
c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.
