Answer:
52.2g of KCl would be left
The mass of KCl will remain despite the solution is been heated
Explanation:
When you are heating a solution, just the solvent (In this case, water), will be evaporated and, in theory, the mass of KCl will remain despite the solution is been heated.
Now, the mass of KCl that you can obtain from 350mL of a 2.0M solution will be:
<em>Moles KCl:</em>
350mL = 0.350L * (2.0mol / L) = 0.700 moles
<em>Mass KCl -Molar mass: 74.55g/mol-:</em>
0.700mol * (74.55g/mol) = 52.2g of KCl would be left
Answer:
Oxidizing agent - CrO4^2-
Reducing agent- N2O
Explanation:
Let us look at the equation closely;
CrO4^2- (aq) + 3N2O(g) ------------> Cr^3+ (aq) + 3NO(g) [acidic]
The reduction half equation is;
CrO4^2- (aq) + 3e -------->Cr^3+ (aq)
Oxidation half equation is;
3N2O(g) ------>3 NO(g) +3 e
Note that the oxidizing agent participates in the reduction half equation while the reducing agent participates in the oxidation half equation as seen above.
The conversion of ethanol to ethanoic acid is an oxidation reaction.
The density of He is 1.79 x 10⁻⁴ g/mL
In other words in 1 mL there's 1.79 x 10⁻⁴ g of He.
To fill a volume of 6.3 L the mass of He required
= 1.79 x 10⁻⁴ g/mL * 6300 mL
= 11 277 * 10⁻⁴ g
Therefore mass of He required = 1.1277 g of He
Answer:
The equation is: CuCO3(s) → CuO(s) + CO2(g)
Explanation:
Copper carbonate decomposes at high temperatures, generating the products carbon dioxide (CO2) and copper oxide (II) CuO. In this type of decomposition reaction, a substance is broken generating two different compounds.
