Answer:
120 g of NaCl in 300 g H20 at 90 C
Explanation:
At x = 90 go vertical to the line for NaCl...then go left to the y-axis to find the solubility in 100 g H20 = 40
we want 300 g H20 so multiply this by 3 to get 120 gm of NaCl in 300 g
Answer:
8.4
Explanation:
-log(4.08x10^-9) = 8.4
- Hope that helped! Please let me know if you need further explanation.
The formula for mole fraction is:
-(1)
The solubility of oxygen gas = 1.0 mmol/L (given)
1.0 mmol/L means 1.0 mmol are present in 1 L.
Converting mmol to mol:
So, moles of oxygen = 0.001 mol
For moles of water:
1 L of water = 1000 mL of water
Since, the density of water is 1.0 g/mL.
So, the mass of water is 1000 g.
Molar mass of water = 18 g/mol.
Number of moles of water = 
Substituting the values in formula (1):
Hence, the mole fraction is 
.
Metals :-
Group 1A - Alkali metals ( highly reactive metals)
Non-metals :-
Group 17 - Halogens ( highly reactive non-metals )
Answer:
moles of carbon dioxide produced are 410.9 mol.
Explanation:
Given data:
Mass of C₆H₁₄O₂ = 16.5 g
Moles of O₂ = 499 mol
Moles of CO₂ = ?
First of all we will write the balance chemical equation.
2C₆H₁₄O₂ + 17O₂ → 14CO₂ + 12H₂O
moles of C₆H₁₄O₂ = mass × molar mass
moles of C₆H₁₄O₂ = 16.5 g × 118 g/mol
moles of C₆H₁₄O₂ = 1947 mol
Now we compare the moles of CO₂ with moles of O₂ and C₆H₁₄O₂ from balance chemical equation.
O₂ : CO₂
17 : 14
499 : 14/17× 499 = 410.9 moles
C₆H₁₄O₂ : CO₂
2 : 14
1947 : 14/2× 1947 = 13629 moles
Oxygen will be limiting reactant so moles of carbon dioxide produced are 410.9 mol.
