a 10.99g sample of NaBr contains 22.34% Na by mass. Considering the law of constant composition (define proportions), how many g
1 answer:
Given :
A 10.99 g sample of NaBr contains 22.34% Na by mass.
To Find :
How many grams of sodium does a 9.77g sample of sodium bromine contain.
Solution :
By law of constant composition , in any given chemical compound, the elements always combine in the same proportion with each other.
Therefore , percentage of Na by mass in NaBr will be same for every amount .
Percentage of Na in 9.77 g NaBr is 22.34 % too .
Gram of Na = .
Hence , this is the required solution .
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Answer:
The energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J
Explanation:
Let the charge on the cation be q₁
Also let the charge on the anion be q₂
A cation q₁ with a valence of 1, has a charge of 1 X 1.602×10⁻¹⁹C = 1.602×10⁻¹⁹C
An anion q₂ with a valence of 3, has a charge of 3 X 1.602×10⁻¹⁹C = 4.806 ×10⁻¹⁹C
The distance between the two charges is 7.5nm = 7.5 X10⁻⁹m
Energy of attraction =
Where k is coulomb's constant = 8.99 X 10⁹ Nm₂/C₂
Energy of attraction =
Energy of attraction = 1.231 X 10⁻¹¹ J
Therefore, the energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J