Answer:
Final calculated molarity of the unknown acid be lower than the actual concentration.
Explanation:
Volume required unknown acid = V= 25.0 mL
Volume actually measure was more than V that is = V'
V' > V
Molarity of the solution is the moles of compound in 1 Liter solutions.
As we can see from the formula that molarity is inversely proportional to the volume of the solution.
- Molarity of solution decreases with increase in volume
- Molarity of solution Increases with decrease in volume.
We have added more volume than required which will be increase the volume of solution and the molarity of the solution will lower than the actual value.
Answer:
The answer is 28 gram.
28 gram of Nitrogen gas are found in 22.4 l in Nitrogen gas
Answer:
Explanation:
Ideal gases follow the combined law of gases:
Where,
- Pressure is the absolute pressure and its units may be in any system, as long as they are the same for both states.
- Also, volume may be in any units, as long as it they are the same for both states.
- Temperature must be absolute temperature, whose unit is Kelvin.
Your data are:
- P₁ = 1200.00 mmHg
- P₂ = 1.11842 atm
- V₁ = 85.0 mL
- V₂ = 350.0 mL
- T₂ = ?
- T₁ = 90.0ºC
<u>1. Conversion of units:</u>
- P₁ = 1200.00 mmHg × 1.00000 atm / 760.000 = 1.578947 mmHg
- T₁ = 90.00ºC + 273.15 = 363.15K
<u>2. Solution</u>
- Clearing T₂, from the combined gas equation you get:
Answer: Stochiometric coefficients
Explanation: The chemical equations must be balanced to follow the law of conservation of mass which says that the mass of products must be same as the mass of reactants.
This is a skeletal equation and needs to be balanced by adjusting the stochiometric coefficients. In order to keep the mass same, the number of atoms must be same on both sides.
Answer:
Explanation:
Ba(s) + Mn²⁺ (aq,1M) → Ba²⁺ (aq,1M) + Mn(s)
Ba⁺²(aq) +2e → Ba(s) , E° = −2.90 V
Mn⁺²(aq) +2e → Mn(s), E⁰ =0.80 V
Anode reaction :
Ba(s) → Ba⁺²(aq) +2e E° = −2.90 V
Cathode reaction :
Mn⁺²(aq) +2e → Mn(s) E⁰ =0.80 V
Cell potential = Ecathode - Eanode
Ecell = .80 - ( - 2.90 )
Ecell = 3.7 V .
equilibrium constant ( K ) :
Ecell = .059 log K / n
n = 2
3.7 = .059 log K / 2
log K = 125.42
K = 2.63 x 10¹²⁵ .
Free energy change :
ΔG = - n F Ecell
= - 2 x 96500 x 3.7
= 714100 J
= 7.141 x 10⁵ J .