In an oxidation-reduction reaction there is an exchange of electrons.
The exchange of electrons implies change in the oxidation states: at least one element increases its oxidation number while other reduces it.
By simple ispection you can predict that in the equation b. there is a change in oxidation states of Cl and Mn.
Now you can check it:
Equation 4H Cl + Mn O2 -> Mn Cl2 + 2H2 O + Cl2
oxidation sates 1+ 1- 4+ 2- 2+ 1- 1+ 2- 0
The oxidation state of Cl in HCl is 1- and it changed to 0 in Cl2
The oxidation state of Mn in MnO2 is 4+ and it changed to 2+ in MnCl2
Answer b.
Answer: 400K
Explanation:
Given that,
Original volume of balloon V1 = 3.0L
Original temperature of balloonT1 = 27°C
Convert the temperature in Celsius to Kelvin
(27°C + 273 = 300K)
New volume of balloon V2 = 4.0L
New temperature of balloon T2 = ?
Since volume and temperature are given while pressure is constant, apply the formula for Charle's law
V1/T1 = V2/T2
3.0L/300K = 4.0L/T2
To get the value of T2, cross multiply
3.0L x T2 = 4.0L x 300K
3.0LT2 = 1200LK
Divide both sides by 3.0L
3.0LT2/3.0L = 1200LK/3.0L
T2 = 400K
Thus, at a temperature of 400 Kelvin, the balloon would have a volume of 4.0L.
Celsius: -11.7
Kelvin: 261.5
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Answer:
14.3 g SO₃
Explanation:
2S + 3O₂ → 2SO₃
First, find the limiting reactant. To do that, calculate the mass of oxygen needed to react with all the sulfur.
5.71 g S × (1 mol S / 32 g S) = 0.178 mol S
0.178 mol S × (3 mol O₂ / 2 mol S) = 0.268 mol O₂
0.268 mol O₂ × (32 g O₂ / mol O₂) = 8.57 g O₂
There are 10.0 g of O₂, so there's enough oxygen. The limiting reactant is therefore sulfur.
Use the mass of sulfur to calculate the mass of sulfur trioxide.
5.71 g S × (1 mol S / 32 g S) = 0.178 mol S
0.178 mol S × (2 mol SO₃ / 2 mol S) = 0.178 mol SO₃
0.178 mol SO₃ × (80 g SO₃ / mol SO₃) = 14.3 g SO₃