Answer:
Explanation:
The graduated cylinder contains 
of water
mL is a volume unit.
Water volume = 41.7 mL
The lead ball caused an increase of volume from 41.7 mL to 96.0 mL
The new volume is the lead ball volume plus the original water volume :
Final volume = Vlead ball+ Water original volume
This is actually true if we suppose that the lead ball is fully sunken in the water.
We always must consider that the volume difference is the volume that the sunken object is occupying in the water.
Answer:
A
Explanation:
In a galvanic cell, energy is produced by spontaneous chemical processes.
The cathode and anode of this cell will depend on the relative position of the two metals in the electrochemical series.
Aluminium is higher in the electrochemical series so aluminium will be the anode. Silver is lower in the electrochemical series so silver will be the cathode.
Recall that oxidation (electron loss) occurs at the anode while reduction (electron gain) occurs at the cathode.
The question is incomplete, the complete question is;
When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate PbSO₄ is reduced to lead at the cathode and oxidized to solid lead(II) oxide PbO at the anode.
Suppose a current of 96.0 A is fed into a car battery for 37.0 seconds. Calculate the mass of lead deposited on the cathode of the battery. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
Answer:
3.81 g of lead
Explanation:
The equation of the reaction is;
Pb^2+(aq) + 2e ---->Pb(s)
Quantity of charge = 96.0 A * 37.0 seconds = 3552 C
Now we have that 1F = 96500 C so;
207 g of lead is deposited by 2 * 96500 C
x g of lead is deposited by 3552 C
x = 207 * 3552/2 * 96500
x = 735264/193000
x = 3.81 g of lead
The amount remaining at the end of 5 half-lives is 7.81×10¹³ g
From the question given above, the following data were obtained:
- Half-life (t½) = 5730 years
- Original amount (N₀) = 2.5×10¹⁵ g
- Number of half-lives (n) = 5
- Amount remaining (N) =?
The amount remaining can be obtained as follow:
N = 1/2ⁿ × N₀
N = 1/2⁵ × 2.5×10¹⁵
N = 1/32 × 2.5×10¹⁵
N = 0.03125 × 2.5×10¹⁵
N = 7.81×10¹³ g
Therefore, the amount remaining after 5 half-lives is 7.81×10¹³ g
Learn more about half-life: brainly.com/question/25783920
