Answer:
The correct answer is "11.44 ml".
Explanation:
Molarity,
= 0.2 M
Density,
= 1.05 g/ml
Volume,
= 1 L
As we know,
⇒ 
or,
⇒ 
On putting the values, we get
⇒ 
⇒ 
Now,
⇒ 
or,
⇒ 
⇒ 
⇒ 
hence,
⇒ 
or,
⇒ 
⇒ 
⇒ 
Answer:
V = 10.3 L
Explanation:
Given data:
Mass of methane = 6.40 g
Volume of CO₂ produced = ?
Temperature = 35°C (35+273 = 308 K)
Pressure = 100.0 KPa (100.0/101 = 0.98 atm)
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Number of moles of CH₄:
Number of moles = mass/molar mass
Number of moles = 6.40 g/ 16 g/mol
Number of moles = 0.4 mol
Now we will compare the moles of CO₂ with CH₄.
CH₄ : CO₂
1 : 1
0.4 : 0.4
Volume of CO₂:
Formula:
PV = nRT
0.98 atm ×V = 0.4 mol ×0.0821 atm.L/mol.K × 308 K
0.98 atm ×V = 10.11 atm.L
V = 10.11 atm.L /0.98 atm
V = 10.3 L
Answer:
The rain containing excess of acid I called acid rain
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:
C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O
Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.
1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C
Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H
The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%
2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:
Amount of C = 0.01138 mol
Amount of H = 0.014244 mol
Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25
The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5
Thus, the empirical formula of the hydrocarbon is C₄H₅.
3. The molar mass of the empirical formula is
Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.
Molecular Formula = C₈H₁₀
Answer:most of the positively charge particles should be bounce back at a range of angles as they collide with the atoms in the foil; only a few should pass straight through the foil
Explanation:
