Answer:
6.58 atm total
3.29 atm NO2
3.29 atm O2
Explanation:
Balanced equation:
O3 + NO → NO2 + O2
There are equal numbers of moles of both reactants, so neither is in excess and either could be considered the limiting reactant.
( 1.30 mol NO) x (1 mol NO2 / 1 mol NO) = 1.30 mol NO2
( 1.30 mol NO) x (1 mol O2 / 1 mol NO) = 1.30 mol O2
<em>Total pressure by using the formula;</em>
P = nRT / V
= ( 1.30 mol + 1.30 mol) x (0.08205746 L atm/K mol) x (401.0 K) / (13.0 L)
= 6.58 atm
<em>Partial pressure for NO2;</em>
(6.58 atm) x (1.30 mol NO2) / (1.30 mol + 1.30 mol)
= 3.29 atm NO2
<em>Partial pressure for O2</em>
6.58 atm total - 3.29 atm NO2
= 3.29 atm O2
Answer:
25 mL
Explanation:
Step 1: Given data
- Concentration of the concentrated solution (C₁): 2 M
- Volume of the concentrated solution (V₁): ?
- Concentration of the diluted solution (C₂): 0.1 M
- Volume of the diluted solution (V₂): 0.500 L
Step 2: Calculate the volume of the concentrated NaCl solution
We will use the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.1 M × 0.500 L / 2 M
V₁ = 0.025 L = 25 mL
13 is the number for it on the periodic table and Group 3 is the correct answer hope this helps.