you are given the number of moles, from the given formula you can get the molar mass which is 12.00g/mol + 16 g/mol =28 g/mol. using the formula n=m/Mr rearrange the formula and make m subject, you thn have m=nMr therefore 7.60x 1021 mol x 28g/mol= m
m = 217268.8 g, you will notice that mol cancel out each other leaving you with g.
now convert g to litres
1 litre is equal to 1000 grams therefore
217268.8g x 1l/1000g = 217,269 litres
<u>Answer:</u> The molality of barium chloride solution is 0.289 m
<u>Explanation:</u>
To calculate the mass of water, we use the equation:
Density of water = 1 g/mL
Volume of water = 748 mL
Putting values in above equation, we get:
To calculate the molality of solution, we use the equation:
where,
= Given mass of solute (barium chloride) = 45.0 g
= Molar mass of solute (barium chloride) = 208.23 g/mol
= Mass of solvent (water) = 748 g
Putting values in above equation, we get:
Hence, the molality of barium chloride solution is 0.289 m
Answer:
to make a discovery, test a hypothesis, or demonstrate a known fact.
Explanation:
perform a scientific procedure, especially in a laboratory, to determine something.
Answer:
C) MnO₂
Explanation:
If we write the semi equations (this is, for reduction and oxidation) we will have:
Oxidation: 2MnO₂ → MnO₄⁻ + Mn⁺²
Reduction: 2H₂O → H₂ + 2OH⁻
Then the final global reaction (just adding the two above hemi chemical reactions):
2MnO₂ + 2H₂O → MnO⁻₄ + Mn⁺² + H₂ + 2OH⁻ (in acidic media)
So, we have Mn⁺⁴ (from MnO₂) goes to Mn⁺⁷ and Mn⁺² (this is, Mn is oxidated) and H⁺ (from acidic media or H₂O) goes to H₂ (this is, it is reduced). So the reducing agent would be MnO₂
Answer:
28
Explanation:
CnH2n+2 is the general molecular forma for a acyclic alkane
