Answer:
220.42098 amu
Explanation:
(220 .9 X .7422) + (220 X .0.1278) + (218.1 X 0.13) = 220.42098 amu
These are weighted averages.
So, we will take mass of one and multiply by abundance percentage that is provided and add them together.
In order to calculate the average atomic mass, we have to convert the percentages of abundance to decimals. So, you get
(220 .9 X .7422) + (220 X .0.1278) + (218.1 X 0.13) = 220.42098 amu
Answer:
P4 + 5O2 --> 2P2O5
2C2H6 + 7O2 --> 4CO2 + 6H2O
Explanation:
We are going to use this formula:
Δ H = T * ΔS
when Δ H is the change in enthalpy
and T is the temperature in Kelvin = 100+273= 373 K
and ΔS is the change in entropy
but first, we need to get the value of ΔH:
ΔH = mass * molar mass per mole * enthalpy of vaporization
by substitution:
ΔH = 51.1 g * (1 mole/18g) * 40.67KJ/mole
= -115.5 KJ
we use negative sign because this is an exothermic reaction.
by substitution on the first formula, we will get the change of entropyΔS:
ΔS = ΔH / T
= -115.5 KJ / 373K
= 310 J/K
MM Zn(NO₃)₂ = 189.36 g/mol
mass 1 mol Zn(NO₃)₂ = 189.36 g
mass hydrate = 100 / 63.67 x 189.36 = 297.409 g
mass 1 mol hydrate = 297.409 g
MM hydrate = 297.409 g/mol
MM hydrate = MM Zn(NO₃)₂ + MM xH₂O
297.409 = 189.36 + x(18)
x = 6
Answer:
468 h
Explanation:
Let's consider the reduction of chromium (III) to chromium that occurs in the electrolytic purification.
Cr³⁺ + 3 e⁻ → Cr
We can establish the following relations.
- 1 kg = 1,000 g
- The molar mass of Cr is 52.00 g/mol
- 1 mole of Cr is deposited when 3 moles of electrons circulate
- The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
- 1 A = 1 c/s
- 1 h = 3,600 s
The hours that will take to plate 11.5 kg of chromium onto the cathode if the current passed through the cell is held constant at 38.0 A is:
