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Flura [38]
3 years ago
13

Seawater is typically 3.5% salt and has a density of 1.03 g/mL. How many grams of salt would be needed to prepare enough seawate

r solution to completely fill a 62.5 L tan
Chemistry
1 answer:
Bas_tet [7]3 years ago
8 0

Answer:

Amount of salt needed is around 2.3*10³ g

Explanation:

The salt content in sea water = 3.5 %

This implies that there is 3.5 g salt in 100 g sea water

Density of seawater = 1.03 g/ml

Volume of seawater = volume of tank = 62.5 L = 62500 ml

Therefore, the amount of seawater required is:

=Density*Volume = 1.03g/ml*62500ml = 6.44*10^{4} g

The amount of salt needed for the calculated amount of seawater is:

=\frac{6.44*10^{4}g\ water*3.5g\ salt }{100g\ water} =2254 g =2.3*10^{3} g

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The percentage error is given by multiplying relative error by 100%.
Therefore, to get the percentage error we need relative error which is given by dividing the absolute error with the actual value. 
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The percentage error is therefore; 2.38%
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3 years ago
A horizontal row of elements in the periodic table represents a<br> group<br> period
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Answer:

Period

Explanation:

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The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.The electron affinity also increases along period because of greater positive charge.

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3 years ago
How is a lithium atom (Li) different from a lithium ion (Li+)?
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7 0
3 years ago
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