Temperature of 62.0 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium
this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.
c1v1 = c2v2
where c1 is concentration and v1 is volume of the concentrated solution
and c2 is concentration and v2 is volume of the diluted solution to be prepared
50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL
substituting these values in the formula
1.50 M x 50.0 mL = C x 250 mL
C = 0.300 M
concentration of the final solution is A) 0.300 M

Here we go ~
1 mole of
has 6.022 × 10²³ molecules of the given compound.
So, 0.78 mole of
will have ~


Explanation:
It is known that one mole of chromium or molar mass of chromium is 51.99 g/mol.
It is given that number of moles is 11.9 moles.
Therefore, calculate the mass of chromium in grams as follows.
No. of moles = 
mass in grams = No. of moles × Molar mass
= 11.9 moles × 51.99 g/mol
= 618.68 g
Thus, we can conclude that there are 618.68 g in 11.9 moles of chromium.