Answer:
3. 116.5 V
4. 119.6 V
Explanation:
3. Determination of the voltage.
Resistance (R) = 25 Ω
Current (I) = 4.66 A
Voltage (V) =?
V = IR
V = 4.66 × 25
V = 116.5 V
Thus, the voltage is 116.5 V
4. Determination of the voltage.
Current (I) = 9.80 A
Resistance (R) = 12.2 Ω
Voltage (V) =?
V = IR
V = 9.80 × 12.2
V = 119.6 V
Thus, the voltage is 119.6 V
Answer:
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fortnite redeem code
Correct Question: what is the oxidizing agent in the reaction.
2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)
Answer: MnO4-is the oxidizing agent
Explanation:
In the reaction 2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)
Oxidizing agent oxidizes other molecules while the themselves get reduced.
oxidizing agents give away Oxygen to other compounds.
MnO4-is the oxidizing agent because
On the reactants side
Oxidation number of Mn in 2MnO4- is +7
Oxidation number of Cl- is -1
On the products side
Oxidation number of Mn is +2
While oxidation number of Cl is zero
Therefore the oxidizing agent is 2MnO4 because is oxidizes Chlorine from -1 to 0 while itself got reduced from oxidation state of +7 to +2
Answer:
Frequency = 6.16 ×10¹⁴ Hz
λ = 4.87×10² nm
Explanation:
In case of hydrogen atom energy associated with nth state is,
En = -13.6/n²
For n = 2
E₂ = -13.6 / 2²
E₂ = -13.6/4
E₂ = -3.4 ev
Kinetic energy of electron = -E₂ = 3.4 ev
For n = 4
E₄ = -13.6 / 4²
E₄ = -13.6/16
E₄ = -0.85 ev
Kinetic energy of electron = -E₄ = 0.85 ev
Wavelength of radiation emitted:
E = hc/λ = E₄ - E₂
hc/λ = E₄ - E₂
by putting values,
6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev - (-3.4ev )
6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev
λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s /2.55ev
λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s /2.55× 1.6×10⁻¹⁹ J
λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J
λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J
λ = 4.87×10⁻⁷ m
m to nm:
4.87×10⁻⁷ m ×10⁹nm/1 m
4.87×10² nm
Frequency:
Frequency = speed of electron / wavelength
by putting values,
Frequency = 3×10⁸m/s /4.87×10⁻⁷ m
Frequency = 6.16 ×10¹⁴ s⁻¹
s⁻¹ = Hz
Frequency = 6.16 ×10¹⁴ Hz
Both figures are mixtures,
Figure II is a heterogenous mixture
Figure I is a homogenous mixture
