Answer:
SO₄²⁻(aq) +Sn²⁺(aq) +4H⁺ → H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O
Explanation:
At first calculate the oxidation state of that element which undergoes oxidation as well as reduction.
for SO₄²⁻ the oxidation state of sulphur is +6 and H₂SO₃ the oxidation state of sulphur is +4
So balance equation is
(Reduction) SO₄²⁻ + 4H⁺+ 2e⁻ → H₂SO₃ + H₂O.........................................(1)
(oxidation) Sn²⁺ → Sn⁴⁺ + 2e⁻ .............................................................(2)
Adding equation 1 & 2
we get
SO₄²⁻(aq) +Sn²⁺(aq) +4H⁺ → H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O
The partial pressure of 0.50 Ne gas is 214.71 torr
calculation
the partial pressure of Ne = moles of Ne/total moles x final pressure
find the total moles of the air mixture
that is moles of Ne + moles of K= 0.50 + 1.20 = 1.70 moles
The partial pressure is therefore = 0.50 /1.70 x 730 = 214.71 torr
Answer:
T2 = 550K
Explanation:
From Charles law;
V1/T1 = V2/T2
Where;
V1 is initial volume
V2 is final volume
T1 is initial temperature
T2 is final temperature
We are given;
V1 = 20 mL
V2 = 55 mL
T1 = 200 K
Thus from V1/T1 = V2/T2, making T2 the subject;
T2 = (V2 × T1)/V1
T2 = (55 × 200)/20
T2 = 550K
Answer:
After Eris was discovered, they had to decide whether Eris was a planet or not. If they decided it wasn't a planet, they had to also decide whether Pluto should be counted as a planet since Eris and Pluto were quite similar. They were the same size, and they were both part of the Kuiper Belt.
Explanation:
Answer:
The solution is given below
Explanation:
Heat, q= mc∆T
q= 125g x 4.18 J/g∙°C x (21.18x- 24.28) °C
q= -1619.75J
NEGATIVE SIGN INDICATES THAT HEAT IS ABSORBED.
Enthalpy Change, ∆H = 1619.75 7/ 10.5 g
= 154.26 J/g
No. of moles of KBr = Mass of KBr/ Molecular Weight of KBr
=10.5g/119gmol-1
=0.088 mol
∆H= 1619.75 J/ 0.088 mol
= 18.41 kJ/mol