Mass = moles x molar mass
so mass of 6 moles of h2 is: 6×1×2 = 12g
- Standard reduction potential of Ag/Ag⁺ is 0.80 v and that of Cu⁺²(aq)/Cu⁰ is +0.34 V.
- The couple with a greater value of standard reduction potential will oxidize the reduced form of the other couple.
Ag⁺ will be reduced to Ag(s) and Cu⁰ will be oxidized to Cu²⁺
Anode reaction: Cu⁰(s) → Cu²⁺ + 2 e⁻ E⁰ = +0.34 V
Cathode reaction: Ag⁺(aq) + e → Ag(s) E⁰ = +0.80 V
Cell reaction: Cu⁰(s) + 2 Ag⁺(aq) → Cu⁺²(aq) + 2 Ag⁰(s)
E⁰ cell = E⁰ cathode + E⁰ anode
= 0.80 + (-0.34) = + 0.46 V
<span>C. C4H8
Given that the number of moles of CO2 and H2O produced from the combustion are equal, that means for every carbon atom, there are 2 hydrogen atoms because CO2 has only 1 carbon atom and H2O has 2 hydrogen atoms. So let's look at the available choices and see which one is correct.
A. C2H2
This is a 1 to 1 ratio of carbon to hydrogen. Wrong answer.
B. C2H6
This is a 1 to 3 ratio of carbon to hydrogen. Wrong answer.
C. C4H8
This is a 1 to 2 ratio of carbon to hydrogen. Correct answer.
D. C6H6
This is a 1 to 1 ratio of carbon to hydrogen. Wrong answer.</span>
Answer: 159 grams
Explanation:
Copper (ii) oxide has the chemical formula CuO.
Now given that:
Mass of CuO in grams = ? (let unknown value be Z)
Number of moles = 2.00 moles
Molar mass of CuO = ?
For the molar mass of CuO: Atomic mass of Copper = 63.5g ; Oxygen = 16g
= 63.5g + 16g
= 79.5 g/mol
Apply the formula:
Number of molecules = (mass in grams/molar mass)
2.00 moles = (Z / 79.5 g/mol)
Z = 79.5 g/mol x 2.00 moles
Z = 159g
Thus, there are 159 grams in 2.00 moles of copper (ii) oxide
Answer:
What that means is that when pressure and number of moles are kept constant, increasing the temperature will result in an increase in volume. Likewise, a decrease in temperature will result in a decrease in volume. In your case, the volume of the gas decreased by a factor of about 3, from "140.0 mL" to "50.0 mL".
