Answer:
Check the explanation
Explanation:
When,
pH = -log[H+] = 3.30
[H+] = 

![alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6](https://tex.z-dn.net/?f=alpha%5BY%5E-4%5D%20%3D%20%5BH%2B%5D%5E6%20%2B%20Ka1%5BH%2B%5D%5E5%20%2B%20Ka1Ka2%5BH%2B%5D%5E4%20%2B%20Ka1Ka2Ka3%5BH%2B%5D%5E3%20%2B%20Ka1Ka2Ka3Ka4%5BH%2B%5D%5E2%20%2B%20Ka1Ka2Ka3Ka4Ka5%5BH%2B%5D%20%2B%20Ka1Ka2Ka3Ka4Ka5Ka6)
= 
= 
When,
pH = -log[H+] = 10.15
[H+] = 
Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 =
; Ka6 = 
= 
= 
Mass of electrons are not included when calculating the atomic mass of element. Atomic mass of the element is equal to proton + neurons. Example the the mass number of Nitrogen can be calculated as 7 protons + 7 neutrons to give 14 . thus nitrogen has a mass number of 14
The resistance of the heating element is 21.61 Ω
Given
The power dissipated = 1500 W
Voltage = 180 V
We know that
Power = Voltage * Current
⇒ Power / Voltage = Current
⇒ 1500 W/180 V = Current
⇒ 8.33 A = Current
In order to calculate the resistance of the heating element. We Have to apply the formula
Power = (Current)^2 * Resistance
⇒ Resistance = Power / (Current)^2
⇒ Resistance = 1500 W/ (8.33) ^2
⇒ Resistance = 21.61 Ω
Hence the resistance of the heating element is 21.61 Ω
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Answer:
Explanation:
Cu²⁺ + 2e⁻ → Cu ( copper gets reduced )
Cu → Cu²⁺ + 2e⁻ ( copper gets oxidized )
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
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.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
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:
The temperature is 288, 88K
Explanation:
We use the formula PV= nRT:
T= PV/nR
T= 0,987 atm x12 L/0,50 mol x 0,082 l atm/K mol
<em>T=288,88K</em>