<h3>
Answer:</h3>
2.125 g
<h3>
Explanation:</h3>
We have;
- Mass of NaBr sample is 11.97 g
- % composition by mass of Na in the sample is 22.34%
We are required to determine the mass of 9.51 g of a NaBr sample.
- Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.
In this case,
- A sample of 11.97 g of NaBr contains 22.34% of Na by mass
A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass
% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100
Mass of the element = (% composition of an element × mass of the compound) ÷ 100
Therefore;
Mass of sodium = (22.34% × 9.51 g) ÷ 100
= 2.125 g
Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g
We are given the compounds:
0.2 mol pb(ch3coo)2, 0.1 mol na2s, and 0.1 mol cacl2
Pb will react with both S and Cl to form a solid compound,
therefore:
Pb+2 + S-2 ---> PbS(s) <span>
Pb+2 + 2Cl- ---> PbCl2(s) </span><span>
<span>Lead(II) sulfide and Lead(II) chloride will precipitate.</span></span>
Answer:
<u>Updrafts</u> (and can cause thunderstorms and other violent weather.)
Good luck, hope this helped!
