Molarity = mol/liter
0.708M = 0.098mol/L
Rearrange to find L:
0.098mol/0.708M = .138L
For every liter there is 1000 mL:
.138L • 1000mL =138mL KOH
When 0.514 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.8 C to 29.4 C. Find ⌂E rxn for the combustion of biphenyl in kJ/mol biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/ C.
<span>The answer is - 6.30 * 10^3 kJ/mol
</span>
3.0 × 10¹¹ RBC's (or) 3E11 RBC's
Solution:
Step 1: Convert mm³ into L;
As,
1 mm³ = 1.0 × 10⁻⁶ Liters
So,
0.1 mm³ = X Liters
Solving for X,
X = (0.1 mm³ × 1.0 × 10⁻⁶ Liters) ÷ 1 mm³
X = 1.0 × 10⁻⁷ Liters
Step 2: Calculate No. of RBC's in 5 Liter Blood:
As given
1.0 × 10⁻⁷ Liters Blood contains = 6000 RBC's
So,
5.0 Liters of Blood will contain = X RBC's
Solving for X,
X = (5.0 Liters × 6000 RBC's) ÷ 1.0 × 10⁻⁷ Liters
X = 3.0 × 10¹¹ RBC's
Or,
X = 3E11 RBC's
Answer: The reaction of 1 mol of C to form carbon monoxide in the reaction 2 C(s) + O2(g) ( 2 CO(g) releases 113 kJ of heat. How much heat will be released by the combustion of 100 g of C according the the above information? According to the balanced chemical equation;
Explanation:
