Answer:
5
Explanation:
Given that the formula is;
1/λ= R(1/nf^2 - 1/ni^2)
λ = 93.7 nm or 93.7 * 10^-9 m
R= 1.097 * 10^7 m-1
nf = ?
ni = 1
From;
ΔE = hc/λ
ΔE = 6.63 * 10^-34 * 3* 10^8/93.7 * 10^-9
ΔE = 21 * 10^-19 J
ΔE = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)
21 * 10^-19 J = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)
21 * 10^-19/-2.18 * 10^-18 = (1/nf^2 - 1/1^2)
-0.963 = (1/nf^2 - 1)
-0.963 + 1 = 1/nf^2
0.037 = 1/nf^2
nf^2 = (0.037)^-1
nf^2 = 27
nf = 5
Answer:
= 82%
Explanation:
Percentage purity is calculated by the formula;
% purity = (mass of pure chemical/total mass of sample) × 100
In this case;
1 mole of Ca(NO3)2 = 164 g
but; 164 g of Ca(NO3)2 = 40 g Ca
Therefore; mass of Ca(NO3)2 = 164 /40
= 4.1 g
Thus;
% purity of Ca(NO3)2 = (Mass of Ca(NO3)2/ mass of the sample)× 100
= (4.1 g/ 5 g) × 100
= 82%
Your answer would be a change in odor! Hope this helps! ;D
Hello!
To find the amount of atoms that are in 80.45 grams of magnesium, we will need to know Avogadro's number and the mass of one mole of magnesium.
Avogadro's number is 6.02 x 10^23 atoms, and one mole of magnesium is equal to 24.31 grams.
1. Divide by one mole of magnesium
80.45 / 24.31 = 3.309 moles (rounded to the number of sigfigs)
2. Multiply moles by Avogadro's number
3.309 x (6.02 x 10^23) = 1.99 x 10^24 (rounded to the number of sigfigs)
Therefore, there are 1.99 x 10^24 atoms in 80.45 grams of magnesium.