Answer:
B.How much space an object takes up
6.4 * 6.02 * 10^23 = 3.8528*10^24 atoms
Don't let the fact that it's vanadium throw you off, avagadros constant stays the same for all elements
Answer:
Balanced chemical equation:
AgCl + 2KCN → K[Ag(CN)₂] (aq) + KCl (aq)
Balanced chemical equation:
2Al + 6NaOH →2 Na₃AlO₃ + 3H₂
Explanation:
Chemical equation:
AgCl + KCN → K[Ag(CN)₂] (aq) + KCl (aq)
Balanced chemical equation:
AgCl + 2KCN → K[Ag(CN)₂] (aq) + KCl (aq)
Net ionic equation can not be written because all are present in aqueous form and no precipitation occur.
Chemical equation:
Al + NaOH → Na₃AlO₃ + H₂
Balanced chemical equation:
2Al + 6NaOH →2 Na₃AlO₃ + 3H₂
This is oxidation reduction reaction.
Hydrogen is reduced in this reaction. while aluminium is oxidized.
Answer:
Mg(s) +<em> 2</em> HCl (aq) → H₂(g) + MgCl₂
0.415g of H₂(g) <em>-Assuming mass of Mg(s) = 10.0g-</em>
Explanation:
Balancing the reaction:
Mg(s) + HCl (aq) → H₂(g) + MgCl₂
There are in products two atoms of H and Cl, the balancing equation is:
Mg(s) +<em> 2</em> HCl (aq) → H₂(g) + MgCl₂
<em>Assuming you add 10g of Mg(s) -Limiting reactant-</em>
<em />
10g of Mg are (Atomic mass: 24.305g/mol):
10g × (1 mol / 24.305g) = <em>0.411 moles of Mg</em>
<em>-Theoretical yield is the amount of product you would have after a chemical reaction occurs completely-</em>
Assuming theoretical yield, as 1 mole of Mg(s) produce 1 mole of H₂(g), theoretical yield of H₂(g) is 0.411moles H₂(g). In grams:
0.411mol H₂(g) × (1.01g / mol) = <em>0.415g of H₂(g)</em>