Answer:
6.25 mL
Explanation:
1.25% of 500 mL is ...
0.0125×(500 mL) = 6.25 mL
Since 1.25% of the 500 mL of solution is bleach, that's how much you need. That amount is 6.25 mL.
Answer:
The molar mass is: 18.02 g/mol.
Explanation:
- Mass of two moles of Hydrogen atoms (H2) = 2x 1 g/mol = 2 g/mol.
- Mass of one mole of water (H2O) = 2 g/mol + 16 g/mol = 18 g/mol.
1 mole of Hydrogen= 1.01, so if we have 2 moles of it here, that would be 2.02.
1 mole of Oxygen (that's all we have here)= 16.00
Once you add the two together (2.02+16.00), you will get 18.02.
I hope this made sense! Have a great day!
Answer:
0.641 moles of ethane
Explanation:
Based on the equation:
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)
We can determine ΔH of reaction using Hess's law. For this equation:
<em>Hess's law: ΔH products - ΔH reactants</em>
ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}
<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>
<em />
ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}
ΔH = -1559.7kJ/mol
That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:
1.00x10³kJ * (1mole ethane / 1559.7kJ) =
<h3>0.641 moles of ethane</h3>
First you have a knowledge of bond order which is
B.O=(no. of electrons in bonding orbital - no. of electrons in non-bonding orbital)÷2
Note:
bond strength is directly proportional to bond order.
For oxygen:
B.O=(6-2)/2= 2; after the removal of two electrons(removal occur from non-bonding orbital)
B.O=(6-0)/2= 3 (As B.O increased bond strength increased)
For Nitrogen:
B.O=(6-0)/2= 3; after the removal of two electrons(removal occur from bonding orbital)
B.O=(4-0)/2= 2 (As B.O decreased bond strength decreased)
The concentration of the solution is 5.0 molar, which is 5.0 mole/L. So in the 1.0 L of 5.0 molar KF salt solution, the moles of KF is 5.0molar*1.0L=5.0 mole. The molecular weight of KF is given in the question as 58.10 gram/mole, so the grams of KF is 58.10 gram/mole * 5.0 mole = 290.5 gram.
