Hey there!
The answer is D, Salinity.
Salinity is the concentration of salt in water. Ocean water often has high salinity and this can contribute to things like upwelling and water density- but these all start from salinity.
Hope this helps!
Answer:
Lowkey dont know the language
Explanation:
Hope this helps :) I didn’t know how to write subscripts so I wrote it down on some paper.
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Answer:</h3>
2.125 g
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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
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Pure Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Br and Br,
E.N of Bromine = 2.96
E.N of Bromine = 2.96
________
E.N Difference
0.00 (Non Polar/Pure Covalent)
For N and O,
E.N of Oxygen = 3.44
E.N of Nitrogen = 3.04
________
E.N Difference
0.40 (Non Polar/Pure Covalent)
For P and H,
E.N of Hydrogen = 2.20
E.N of Phosphorous = 2.19
________
E.N Difference 0.01 (Non Polar/Pure Covalent)
For K and O,
E.N of Oxygen = 3.44
E.N of Potassium = 0.82
________
E.N Difference 2.62 (Ionic)
