Suppose that a fully charged lead-acid battery contains 1.50 L of 5.00 M H2SO4. What will be the concentration of H2SO4 in the battery after 2.50 A of current is drawn from the battery for 7.50 hours

Answer : The final concentration of is, 5.47 M

Explanation :

First we have to calculate the charge.

Charge = Current × Time

Charge = 2.50 A × 7.50 hr

Charge = 2.50 A × 7.50 × 3600 s

Charge = 67500 C

Now we have to calculate the moles of electrons.

That means,

Moles of consumed = Moles of electrons = 0.699 mol

Now we have to calculate the initial moles of

Final moles of = 7.5 + 0.699 = 8.199 mol

Now we have to calculate the final concentration of

Therefore, the final concentration of is, 5.47 M

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation:

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