I think O and Cl will form covalent bonds since they are both non-metals and don't have a large enough difference in electronegativities to create an ionic bond.
I hope this helps. Let me know if anything is unclear.
Answer:
2.33 *10^1 g
Explanation:
.595 mol *39.0983 g/1 mol = 23.2634 > 2.33 *10^1 g
The answer is 50 kcal by the formula of 10% - 500kcal x .1 = 50 kcal it is the approximately chemical energy that can contain by the secondary consumers.
1%- 500,000 kcal x .01 = 5,000 kcal for producers.
10%- 5,000 kcal x .1 = 500 kcal for primary consumers.
10%- 500 kcal x .1 = 50 kcal for secondary consumers.
10%- 50 kcal x .1 = 5 kcal for tertiary consumers.
Each trophic level receives less and less energy, so fewer organisms can be supported at higher levels.
10% of energy gets passed to next level in form of chemical bonds and 1 decimal place.
<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