Learn how to write, balance net ionic equations<span> with examples. ... </span>A<span> chemical equation shows the reactant molecules which involve in reaction and products which are formed during the reaction as well as .... Na</span>+(aq<span>) + </span>OH-(aq) + H+(aq) + NO3-(aq) ? ... For example: Aqueous Sodium chloride is to be written as Na+<span> and </span>Cl-<span>.
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1 mol N₂ - 2 mol NH₃
6 mol N₂ - x mol NH₃
x=2×6/1=12 mol
12 mol NH₃
By using this formula of vapor pressure:
Pv(solu)= n Pv(water)
when we have Pv(solu)=231.16 torr & Pv(water)= 233.7 torr
from this formula, we can get n (mole fraction of water) by substitution:
231.16 = n * 233.7
∴ n(mole fraction of water) = 0.99
so mole fraction of solution = 1 - 0.99 = 0.01
when no.of moles of water = mass weight / molar weight
= 365g / 18g/mol = 20 moles
Total moles in solution = moles of water / mole fraction of water
= 20 / 0.99 =20.2
no. of moles of the solution= total moles in solution- moles of water
= 20.2 - 20 = 0.2 moles
when we assumed the mass weight of the solution = 16 g (missing in your question should be given)
∴ molar mass = mass weight of solute / no. of moles of solute
= 16 g / 0.2 mol = 80 g/mol
Answer:
Explanation:
We can use the Ideal Gas Law to calculate the density of the gas.
pV = nRT
n = m/M Substitute for n
pV = (m/M)RT Multiply both sides by M
pVM = mRT Divide both sides by V
pM = (m/V) RT
ρ = m/V Substitute for m/V
pM = ρRT Divide each side by RT
Data:
p = 1.00 bar
M = 49 g/mol
R = 0.083 14 bar·L·K⁻¹mol⁻¹
T = 0 °C = 273.15 K
Calculation:
ρ = (1.00 × 49)/(0.083 14 × 273.15) = 2.2 g/L
The density of the gas is .
Answer:
C: 4
H: 1
Mg: 2
O : 6
Explanation:
You can quickly find out the number of valence electrons by looking at where the element is on the periodic table and referring to the table that is attached.
