Answer:
Explanation:
- For the balanced reaction:
<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).</em>
It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.
- Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:
no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.
- Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:
<em><u>Using cross multiplication:</u></em>
4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.
0.64 mol of Fe is needed to react with → ??? mol of O₂.
∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.
- Finally, we can get the volume of oxygen using the information:
<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>
<em></em>
<em><u>Using cross multiplication:</u></em>
1 mol of O₂ occupies → 22.4 L, at STP conditions.
0.48 mol of O₂ occupies → ??? L.
∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.
The solubility of PbBr₂(s) with the presence of 0.500 M of KBr is
2.64 x 10⁻⁵ M.
Answer:
Ionic
Explanation: Ionic compounds tend to be hard and brittle while covalent compounds tend to be softer and more flexible.
Hope this helped!
Answer : The value of is, 0.34 V
Explanation :
Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.
The oxidation-reduction half cell reaction will be,
Oxidation half reaction:
Reduction half reaction:
Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.
The overall balanced equation of the cell is,
To calculate the of the reaction, we use the equation:
Putting values in above equation, we get:
Hence, the value of is, 0.34 V
Answer:
6. O₂ + Cu —> CuO
7. H₂ + Fe₂O₃ —> H₂O + Fe
8. O₂ + H₂ — > H₂O
9. H₂S + NaOH —> Na₂S + H₂O
10. Al + HCl —> H₂ + AlCl₃
Explanation:
6. Oxygen gas react with solid copper metal to form copper(II) oxide
Oxygen gas => O₂
Copper => Cu
copper(II) oxide => CuO
The equation is:
O₂ + Cu —> CuO
7. hydrogen gas and iron(III) oxide powder react to form liquid water and solid iron power
hydrogen gas => H₂
Iron(III) oxide => Fe₂O₃
Water => H₂O
Iron => Fe
The equation is:
H₂ + Fe₂O₃ —> H₂O + Fe
8. Oxygen gas react with hydrogen gas to form liquid water
Oxygen gas => O₂
hydrogen gas => H₂
Water => H₂O
The equation is:
O₂ + H₂ — > H₂O
9. Hydrogen sulphide gas is bubbled through a sodium hydroxide solution to produce sodium sulphide and liquid water
hydrogen sulphide => H₂S
sodium hydroxide => NaOH
Sodium sulphide => Na₂S
Water => H₂O
The equation is:
H₂S + NaOH —> Na₂S + H₂O
10. Hydrogen gas and aluminum chloride solutions are produced when solid aluminum react with hydrochloric acid
Aluminum => Al
Hydrochloric acid => HCl
hydrogen gas => H₂
Aluminum chloride => AlCl₃
The equation is:
Al + HCl —> H₂ + AlCl₃