The formula that can be applied in this problem is W = Fd
where W is work, F is the force and d is distance. You have 450N and 3m, all
you have to do is to multiply it.
W = Fd
W = (450N) (3m)
W = 1350J
The answer is letter C.
Mass of metal piece is 611 g and volume of graduated cylinder is 25.1 mL. When metal piece is placed in the graduated cylinder water level increases to 56.7 mL. The increase in volume is due to volume of metal piece that gets added to the volume of water.
Thus, volume of metal piece can be calculated by subtracting initial volume from the final one.

Thus, volume of metal piece will be 31.6 mL. The mass of metal piece is given 611 g, density of metal can be calculated as follows:

Therefore, density of metal is 19.33 g/mL.
Ойлголоо, уучлаарай Ойлголоо, уучлаарай /; coo
moles NaOH = c · V = 0.2432 mmol/mL · 24.75 mL = 6.0192 mmol
moles H2SO4 = 6.0192 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 3.0096 mmol
Hence
[H2SO4]= n/V = 3.0096 mmol / 38.94 mL = 0.07729 M
The answer to this question is [H2SO4] = 0.07729 M