Answer:
1. 25 moles water.
2. 41.2 grams of sodium hydroxide.
3. 0.25 grams of sugar.
4. 340.6 grams of ammonia.
5. 4.5x10²³ molecules of sulfur dioxide.
Explanation:
Hello!
In this case, since the mole-mass-particles relationships are studied by considering the Avogadro's number for the formula units and the molar mass for the mass of one mole of substance, we proceed as shown below:
1. Here, we use the Avogadro's number to obtain the moles in the given molecules of water:
2. Here, since the molar mass of NaOH is 40.00 g/mol, we obtain:
3. Here, since the molar mass of C6H12O6 is 180.15 g/mol:
4. Here, since the molar mass of ammonia is 17.03 g/mol:
5. Here, since the molar mass of SO2 is 64.06 g/mol:
Best regards!
I believe Winter is <span>your answer.</span>
The answer is b
Explanation:
The answers would be the mass before a reaction is the same as the mass after a reaction which basically means mass is conserved
Explanation:
This is correct!
Ions that exist in both the reactant and product side of the equation are referred to as spectator ions. Overall, they do not partake in the reaction. If they are present on both sides of the equation, you can cancel them out.
An example is;
Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq) → Na+(aq) + NO3−(aq) + AgCl(s)
The ions; Na+, NO3−(aq) would be cancelled out to give;
Cl−(aq) + Ag+(aq) → AgCl(s)