Answer:
THE MOLARITY IS 2.22 MOL/DM3
Explanation:
The solution formed was as a result of dissolving 37.5 g of Na2S in 217 g of water
Relative molecular mass of Na2S = ( 23* 2 + 32) = 78 g/mol
Molarity in g/dm3 is the amount of the substance dissolved in 1000 g or 1 L of the solvent. So we have;
37.5 g of Na2S = 217 g of water
( 37.5 * 1000 / 217 ) g = 1000 g of water
So, 172.81 g/dm3 of the solution
So therefore, molarity in mol/dm3 = mol in g/dm3 / molar mass
Molarity = 172.81 g/dm3 / 78 g/mol
Molarity = 2.22 mol/dm3
The molarity of the solution is 2.22 mol/dm3
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Hey There!</h2><h2>
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Answer:</h2><h2>_____________________________________</h2><h2>

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London Dispersion force or Van de Waals force is a temporary attractive force which are the weakest and occur between nonpolar noble gases and same charges. This force is weaker because they have more electrons that are farther from the nucleus and are able to move around easier.
Example:
, 
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Dipole force is present between the polar molecules. Polar molecules are those molecules which have slightly negative and slightly positive charge. Dipole-dipole forces are attractive forces between the positive end of one polar molecule and the negative end of another polar molecule.
Example: HCl, HF, 
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It is a special type of dipole force present between polar molecules, it is formed between Hydrogen atom which forms positive ion, and the other negative ion. It results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom such as a N, O, or F atom. The hydrogen bond is one of the strongest intermolecular attractions, but weaker than a covalent or an ionic bond.
Example: Every polar molecule which has hydrogen has hydrogen bonding i.e. 
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Hello! The answer is D
A good note for these is when there are three elements, one being a singular element and another a compound and there’s a single switch, this could show a single-displacement
Have a good day gamer.
Some
of the solutions exhibit
colligative properties. These properties depend on the amount of solute
dissolved in a solvent. These properties include freezing point depression, boiling
point elevation, osmotic pressure and vapor pressure lowering. Calculations
are as follows:
<span>
ΔT(freezing point) = (Kf)mi
3 = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg</span>
Answer:
<u>2.26 </u>
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Explanation:
First you should know what are significant figures,
1 . All non - zero digits are significant.
2 . The zero between two non- zero are significant.Ex 302 has 3 significant figures.
3. The zero before the decimal and before any non-zero digit is non significant. Example : 0.003 has only 1 significant figure.
4. The zero after non zero are non - significant . But the zeros after the decimal point are significant.
300 has only 1 significant figure
300.0 has 4 significant figure
The least number of significant figures present in the number should be there in the final answer of the calculation.
Look at the number having minimum significant digits :
It is 3 (every number has 3- significant figures ) So the answer should also contain 3 - significant figures.

First solve the numerator part


Round off this number to 3 significant figures.
Answer is =<u> 2.26</u>
It has 3 - significant digit since all the digits are non-zero.