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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 atoms that would be expected to be diamagnetic in the ground state is  magnesium

The magnetism of an atom refers to its electronic configuration. A diamagnetic atom is an atom whose electrons are all paired.

A paired electron is an electron that occurs in pairs in its orbital shell.

At their respective ground state, the electronic configuration of the given elements are as follows:

The electronic configuration of magnesium is 1s²2s²2p⁶3s².  As such its a diamagnetic atom.

The electronic configuration of Potassium is 1s²2s²2p⁶3s²3p⁶4s¹. Hence, Potassium has one unpaired electron in its outermost shell.

The electronic configuration of Chlorine is 1s²2s²2p⁶3s²3p⁵. Hence, Chlorine has one unpaired electron in its outermost shell.

The electronic configuration of Cobalt is 1s²2s²2p⁶3s²3p⁶3d⁷4s². Hence, the unpaired electrons of Cobalt in its outermost shell are three.

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2 years ago
If 30 ml of water containing 15 mg of salicylate is extracted with 30 ml of chloroform, 2.59 mg of salicylate remains in the wat
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Answer:

Distribution coefficient: 4.79

Explanation:

Distribution coefficient is the ratio between equilibrium concentration of non-aqueous phase and aqueous phase where both solvents are inmiscible. The equation for the problem is:

Distribution coefficient: Concentration in chloroform / Concentration in Water

<em>Concentration in water: 2.59mg / 30mL = 0.08633mg/mL</em>

<em>Concentration in chloroform: (15mg-2.59mg) / 30mL = 0.4137mg/mL</em>

<em />

Distribution coefficient: 0.4137mg/mL / 0.08633mg/mL

<h3>Distribution coefficient: 4.79</h3>
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Explanation:

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