

ok, now press calculator. i dont have it now.
42cubic centimeter are in the block
volume is 48
Hello!
The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.
We have the following data:
mo (initial mass) = 53.3 mg
m (final mass after time T) = ? (in mg)
x (number of periods elapsed) = ?
P (Half-life) = 10.0 minutes
T (Elapsed time for sample reduction) = 25.9 minutes
Let's find the number of periods elapsed (x), let us see:






Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:




I Hope this helps, greetings ... DexteR! =)
We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:
KOH + H₂SO₄ → H₂O + KHSO₄
If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:
0.025 L x 0.150 mol/L = .00375 mol KOH
0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄
We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:
0.00375 mol / 0.015 L = 0.25 mol/L
The concentration of H₂SO₄ being neutralized is 0.25 M.