First we will calculate the number of moles of Iron:
, where n is the number of moles, m is the mass of iron in the reaction and M is the Atomic weight.
moles of Iron.
The same number of moles of Oxygen will take part in the reaction.
So
where 32 is the Atomical Weight of Oxygen (16 x 2).
=>
g
Volume of each solution : 60 ml 20% and 40 ml 45%
<h3>Further explanation</h3>
Given
20% and 45% acid
100 ml of 30% acid
Required
Volume of each solution
Solution
Molarity from 2 solutions :
Vm Mm = V₁. M₁ + V₂. M₂
m = mixed solution
V = volume
M = molarity
V₁ = x ml
V₂ = (100 - x) ml
Input the value :
100 . 0.3 = x . 0.2 + (100-x) . 0.45
30 = 0.2x+45-0.45x
0.25x=15
x= 60 ml
V₁ = 60 ml
V₂ = 100 - 60 = 40 ml
Answer:
pHe = 3.2 × 10⁻³ atm
pNe = 2.5 × 10⁻³ atm
P = 5.7 × 10⁻³ atm
Explanation:
Given data
Volume = 1.00 L
Temperature = 25°C + 273 = 298 K
mHe = 0.52 mg = 0.52 × 10⁻³ g
mNe = 2.05 mg = 2.05 × 10⁻³ g
The molar mass of He is 4.00 g/mol. The moles of He are:
0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol
We can find the partial pressure of He using the ideal gas equation.
P × V = n × R × T
P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K
P = 3.2 × 10⁻³ atm
The molar mass of Ne is 20.18 g/mol. The moles of Ne are:
2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol
We can find the partial pressure of Ne using the ideal gas equation.
P × V = n × R × T
P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K
P = 2.5 × 10⁻³ atm
The total pressure is the sum of the partial pressures.
P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm
The correct answer would be the first option. Material A having a smaller latent heat of fusion would mean that it will take only less energy to phase change into the liquid phase. Latent of heat of fusion is the amount of energy needed of a substance to phase change from solid to liquid or liquid to solid.
Nitrogen lone pair will act as a base,removing H+ from water leaving behind OH- ion.
Why ?
Because N is a better donor than O.
