32. mercury has an atomic mass of 200.59 amu. calculate the a. mass of 3.0 x 1023 atoms. b. number of atoms in one nanogram of mercury? i needed on this question

a.200.59 amu of mercury contains = 6.02 * 1023 molecules

Mass for 3.0 * 1023 molecules = (3.0 * 1023 * 200.59) /6.02 * 1023

= 100.29 amu

Thus, mass for 3.0 * 1023 molecules is 100.29 amu

b.As, 200.59 g mercury contains = 6.02 * 1023 atoms

1 g mercury contains = (6.02 * 1023 /200) atoms

= 3.01 * 1021 atoms

Thus, 1 ng mercury contains = (3.01 * 1021) / 109

= 3.01 * 1012 atoms

Thus, 3.0 * 1012 atoms present in 1 ng mercury

a) Mass of of mercury is 99.91 g.

b) in 1 nano gram of mercury.

Explanation:

Number of molecules or number of atoms:

atoms or molecules

Mercury has an atomic mass of 200.59 amu = 200.59 g/mol

Number of mercury atoms =

Moles of Mercury :

Mass of 0.4981 moles of mercury:

0.4981 moles × 200.59 amu = 99.91 g

Mass of of mercury is 99.91 g.

b) Mass of mercury = 1 nano gram =

Moles of mercury:

Number of atoms of mercury in 1 nano gram:

:

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"