1) mass composition
N: 30.45%
O: 69.55%
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100.00%
2) molar composition
Divide each element by its atomic mass
N: 30.45 / 14.00 = 2.175 mol
O: 69.55 / 16.00 = 4.346875
4) Find the smallest molar proportion
Divide both by the smaller number
N: 2.175 / 2.175 = 1
O: 4.346875 / 2.175 = 1.999 = 2
5) Empirical formula: NO2
6) mass of the empirical formula
14.00 + 2 * 16.00 = 46.00 g
7) Find the number of moles of the gas using the equation pV = nRT
=> n = pV / RT = (775/760) atm * 0.389 l / (0.0821 atm*l /K*mol * 273.15K)
=> n = 0.01769 moles
8) Find molar mass
molar mass = mass in grams / number of moles = 1.63 g / 0.01769 mol = 92.14 g / mol
9) Find how many times the mass of the empirical formula is contained in the molar mass
92.14 / 46.00 = 2.00
10) Multiply the subscripts of the empirical formula by the number found in the previous step
=> N2O4
Answer: N2O4
If we use mile or something else it will be hard to measure in sky .
Because the mile or meters too short for the space,
Galaxy and stars are too long far away from us if we use mile or meter,
It till be take like miliion year to measure.
For a comparison of the nucleus 5626fe, the density of the nucleus 112 48cd is mathematically given as the same.
n(Cd) / n(Fe)=1
<h3>What is the density of the nucleus 112 48cd?</h3>
Generally, the equation for the density is mathematically given as
d=\frac{A}{4/3}\piR^3
Therefore
n(Cd) / n(Fe) = [A (Cd) / (A Fe) ] * [ R (Fe) / R (Cd)]^3
n(Cd) / n(Fe)= (112 / 56 ) * (1/1.26)3
n(Cd) / n(Fe)=1
In conclusion, The ratio of n(Cd) = n(Fe) is 1, hence same
Read more about density
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You have OH- conc = </span>2.3 ✕ 10−6 m
From the formula, you can observe the ratio of Cu2+ to OH- is 4 : 6 = 2:3
So, for 2.3 ✕ 10−6 m OH-
[Cu2+] =
