Answer:
In the electrolysis of dilute sulfuric acid, which electrolysis in aqueous solution to form hydrogen ions, H⁺, and sulfur IV ions SO₄²⁻ in the presence of H⁺ and OH⁻ ions from the water molecules
At the anode
The anode, positive electrode, attracts the negative OH⁻ and SO₄²⁻ ions where the OH⁻ gives up electrons to form water molecules and oxygen as follows;
4OH⁻ → 2H₂O + O₂ + 4e⁻
At the cathode
The positive H⁺ ions from the water molecules and the acid are attracted to the cathode where they combine with 2 electrons to form hydrogen gas as follows;
2e⁻ + 2H⁺ → H₂ (gas)
Explanation:
Answer:
Something like table salt (NaCl) is a compound because it is made from more than one kind of element (sodium and chlorine), but it is not a molecule because the bond that holds NaCl together is an ionic bond. If you like, you can say that sodium chloride is an ionic compound.
Explanation:
Answer: 5
Explanation: add up all the electrons and it will amount to 23. Arranging by the old model for electronic configuration, we have : 2, 8, 8, 5
The last number being 5 represent its valence electron
Answer:
31.9 °C
Explanation:
The formula for the heat q absorbed by an object is
q = mCΔT where ΔT = (T₂ - T₁)
Data:
q = 12.35 cal
m = 19.75 g
C = 0.125 cal°C⁻¹g⁻¹
T₂ = 37.0 °C
Calculations
(a) Calculate ΔT
q = mCΔT
12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT
12.35 = 2.406ΔT °C⁻¹
ΔT = 12.35/(2.406 °C⁻¹) = 5.13 °C
(b) Calculate T₂
ΔT = T₂ - T₁
T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C
The original temperature was 31.9 °C.
Answer: The correct answer is -297 kJ.
Explanation:
To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:
2SO3 —> O2 + 2SO2 (196 kJ)
Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.
Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:
SO3 —>1/2O2 + SO2 (98 kJ)
S + 3/2O2 —> SO3 (-395 kJ)
Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.
S + O2 —> SO2
Now, we must add the enthalpies together to get our final answer.
-395 kJ + 98 kJ = -297 kJ
Hope this helps!
