I am not all understood but for the school to earn money you can:
make
--a raffle
--lotto
-- yard sale
-- class photo
-- origami for sale or something
-- buffet or food sale (example all Friday ice cream sale, 2 livre ice cream)
Answer:
Cu(NO3)2(aq)+Pb(s) ⇌ Pb(NO3)2(aq)+Cu(s)
Explanation:
If we look at the both reactions closely, we will quickly discover that the reaction CuSO4(aq)+Pb(s) ⇌ PbSO4(s)+Cu(s) involves PbSO4.
The compound PbSO4 is insoluble in water and sinks to the bottom of the reaction vessel. When this occurs, the concentration of Pb^2+ becomes low. This will bring about a low voltage in the cell.
On the other hand, Pb(NO3)2 is soluble in water hence the cell voltage in this case is higher than the former.
Answer:
C₂H₄O
CH₃CHO
Explanation:
I'm not sure if you want the molecular formula or the condensed structure, but I will give you both.
Molecular formula:
You have 2 carbons (C₂), 4 hydrogens (H₄), and 1 oxygen (O). The molecular formula will be C₂H₄O.
Condensed Structure:
You have a carbon bonded to three hydrogens (CH₃). This carbon is bonded to a carbon that is bonded to a hydrogen and oxygen (CHO). The condensed structure will be CH₃CHO.
0.012moldm⁻³
Explanation:
Given parameters:
Mass of AgNO₃ = 1000mg
Volume of water = 500mL
Unknown:
Molarity of solution = ?
Solution:
The molarity of a solution is the number of moles of a solute dissolved in volume of solvent.
Molarity = 
Number of moles of AgNO₃ = ?
Number of moles = 
Molar mass of AgNO₃ = 108 + 14 + 3(16) = 170g/mol
convert mass to g;
1000mg = 1g
Number of moles = 
= 0.00588moles
convert the given volume to dm³;
1000mL = 1dm³;
500mL = 0.5dm³
Now solve;
Molarity = 
= 0.012moldm⁻³
learn more:
Molarity brainly.com/question/9324116
#learnwithBrainly
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
