They must have both cations and anions.
I hope this helps.
Answer:
See Explanation
Explanation:
moles of NH₃ = 11.9g/17.03 g/mol = 0.699 mole
moles of CN₂OH₄ = 1/2(0.699) mole =0.349 mole
Theoretical yield of CN₂OH₄ = (0.349 mole)(60 g/mole) = 20.963 grams
%Yield = Actual Yield/Theoretical Yield x 100%
= 18.5g/20.963g x 100% = 88.25%
3.25 C because I said so and energy is green
<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
