Answer:
c. -1020.9 kJ
Explanation:
4Fe (s) + 3 O₂ (g) --> 2 Fe₂O₃(s) ΔH = -826.0 kJ/mol.
atomic weight of iron = 56
69.03 g = 69.03 / 56
= 1.23268 moles
Heat released by 1.23268 moles
= 1.23268 x 826.0
= -1020.9 kJ .
Answer:
a. glucose in water( solution)
b. smoke in air (colloids)
c. carbon dioxide in air (solution)
d. milk( colloids)
Explanation:
A solution is said to be formed when a solute dissolves in a solvent to form a homogeneous mixture. The solute particles are less than 10^-9m in size. Familiar solutions are those where the solute are dissolved in a liquid solvent. When the liquid water, the solution is known as an aqueous solution. A typical example is (glucose in water). In some other cases, the apparent solution of a solute in a solvent is accompanied by a chemical reaction and this is often known as a chemical reaction. A typical example is (carbon dioxide in air).
Colloids are also known as false solutions. Here, the individual solute particles are larger than the particles of the true solution, but not large enough to be seen by the naked eye. When a light beam is placed beside a beaker containing a colloid, the light rays of the beam can be clearly seen. This shows that it exhibits the Tyndall effect while a solution dosent exhibit such.
In a colloid, the liquid solvent is more appropriately know as the DISPERSION medium while the solid solute particles constitute the DISPERSED substance. This can either be solid, liquid or gas.
For example:
--> smoke in air : Dispersion medium is gas while the dispersed substance is solid.
--> milk: Dispersion medium is liquid while the dispersed substance is liquid.
Data Given:
Time = t = 30.6 s
Current = I = 10 A
Faradays Constant = F = 96500
Chemical equivalent = e = 63.54/2 = 31.77 g
Amount Deposited = W = ?
Solution:
According to Faraday's Law,
W = I t e / F
Putting Values,
W = (10 A × 30.6 s × 31.77 g) ÷ 96500
W = 0.100 g
Result:
0.100 g of Cu²⁺ is deposited.
Answer: 40 grams
Explanation:
The quantity of Heat Energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since Q = 93.4J
M = ?
C = 0.129 J/g.C
Φ = 40.4°C - 22.3°C = 18.1°C
Then, Q = MCΦ
Make Mass, M the subject formula
M = Q/CΦ
M = (93.4J) / (0.129 J/g.C x 18.1°C)
M = 93.4J / 2.33J/g
M = 40 g
Thus, the mass of the lead is 40 grams
