The reaction of ethane burning:
2C2H6 + 7O2 -> 4CO2 + 6H2O
So 2:6 = 3 mol: x
x = 9 mol
A. The new molarity of the diluted solution is 0.047 M
B. The new molarity of the diluted solution is 0.188 M
C. The new molarity of the diluted solution is 4.4 M
D. The new molarity of the diluted solution is 1.28 M
A. How to determine the new molarity
- Volume of stock solution (V₁) = 26.6 mL
- Molarity of stock solution (M₁) = 0.127 M
- Volume of water = 45.4 mL
- Volume of diluted solution (V₂) = 26.6 + 45.4 = 72 mL
- Molarity of diluted solution (M₂) =?
M₁V₁ = M₂V₂
0.127 × 26.6 = M₂ × 72
Divide both side by 72
M₂ = (0.127 × 26.6) / 72
M₂ = 0.047 M
B. How to determine the new molarity
- Volume of stock solution (V₁) = 49.5 mL
- Molarity of stock solution (M₁) = 0.757 M
- Volume of water = 150 mL
- Volume of diluted solution (V₂) = 49.5 + 150 = 199.5 mL
- Molarity of diluted solution (M₂) =?
M₁V₁ = M₂V₂
0.757 ×49.5 = M₂ × 199.5
Divide both side by 199.5
M₂ = (0.757 × 49.5) / 199.5
M₂ = 0.188 M
C. How to determine the new molarity
- Volume of stock solution (V₁) = 102 mL
- Molarity of stock solution (M₁) = 25.8 M
- Volume of water = 500 mL
- Volume of diluted solution (V₂) = 102 + 500 = 602 mL
- Molarity of diluted solution (M₂) =?
M₁V₁ = M₂V₂
25.8 × 102 = M₂ × 602
Divide both side by 602
M₂ = (25.8 × 102) / 602
M₂ = 4.4 M
D. How to determine the new molarity
- Volume of stock solution (V₁) = 741 mL
- Molarity of stock solution (M₁) = 1.81 M
- Volume of water = 308 mL
- Volume of diluted solution (V₂) = 741 + 308 = 1049 mL
- Molarity of diluted solution (M₂) =?
M₁V₁ = M₂V₂
1.81 × 741 = M₂ × 1049
Divide both side by 1049
M₂ = (1.81 × 741) / 1049
M₂ = 1.28 M
Learn more about dilution:
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Answer:
62.5 km/hr
Explanation:
Speed can be found by dividing the distance by the time.
s=d/t
We know the car travelled a distance of 125 kilometers in 2 hours. Therefore:
• d=125 km
• t= 2 hr
Substitute the values into the formula.
s= 125 km/ 2 hr
Divide
s= 62.5 km/ hr
The speed of the car is 62.5 kilometers per hour.
Answer:
3AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
The coefficients are 3, 1, 3, 1
Explanation:
From the question given above, the following data were:
Silver chloride reacts with sodium phosphate to yield sodium chloride and silver phosphate. This can be written as follow:
AgCl + Na₃PO₄ —> NaCl + Ag₃PO₄
The above equation can be balanced as follow:
AgCl + Na₃PO₄ —> NaCl + Ag₃PO₄
There are 3 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 3 in front of NaCl as shown below:
AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
There are 3 atoms of Cl on the right side and 1 atom on the left. It can be balance by putting 3 in front of AgCl as shown below:
3AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
Thus, the equation is balanced.
The coefficients are 3, 1, 3, 1
The original results have not been replicated consistently and reliably.
