Answer:
Reduction
Explanation:
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Mn⁺⁷ +3e⁻ → Mn⁴⁺
Mn gets three electrons , its oxidation state reduced from +7 to +4 so Mn gets reduced.
Examples:
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
Answer:
19.39 mL.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have different values of V and T:
<em>(V₁T₂) = (V₂T₁)</em>
<em></em>
V₁ = 22.8 mL, T₁ = 48°C + 273 = 321 K,
V₂ = ??? mL, T₂ = 273 (standard temperature = 273 K),
- Applying in the above equation
<em>(V₁T₂) = (V₂T₁)</em>
<em />
∴ V₂ = (V₁T₂)/(T₁) = (22.8 mL)(273 K)/(321 K) = 19.39 mL.
Answer:
Ca2+ is an ion with a correct charge.
<em>A glass flask of volume 400 cm³ is just filled with mercury at 0°C. How much mercury will overflow when the temperature of the system rises to 80°C.</em>
<em />
The volume of mercury that overflow is 5.376 cm³
<h3>Further explanation</h3>
Given
volume of glass = 400 cm³
Δt=80 °C - 0 °C = 80
Required
overflow volume
Solution
With an increase in the temperature of the substance, objects can expand. This expansion includes volume expansion.
Can be formulated
Find volume expansion of glass and mercury
Overflow :
ΔV mercury - ΔV glass : 5.76-0.384 = 5.376 cm³
Answer:
Explanation:
Repeated tests help determine whether the hypothesis is always true in different circumstances. Repeatedly testing a hypothesis qualifies it to be accepted as a theory.
