+109 I'm feeling lazy today, so I'm not gonna do some grand write-up of how I got the answer. However I'll give you the basics.
$$Grams \Leftrightarrow Mols \Leftrightarrow Atoms$$
Since you know how many atoms of calcium you have, you'll divide that number by Avogadro's Constant (6.022 x 1023) to deduce the amount of mols of Calcium you have, giving:
$${\mathtt{0.516}}{\mathtt{\,\times\,}}{\mathtt{molCa}}$$
Checking the periodic table shows us that Calcium has a molar weight of 40.08g/mol Ca. Multiply the number of mols, and it's molar weight, giving:
$${\mathtt{0.516}}{\mathtt{\,\times\,}}{\mathtt{molCa}}{\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{40.08}}{\mathtt{\,\times\,}}{\mathtt{gCa}}\right)}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molCa}}\right)}}\right) = {\mathtt{20.68}}{\mathtt{\,\times\,}}{\mathtt{gCa}}$$
Good luck.
+109 I'm feeling lazy today, so I'm not gonna do some grand write-up of how I got the answer. However I'll give you the basics.
$$Grams \Leftrightarrow Mols \Leftrightarrow Atoms$$
Since you know how many atoms of calcium you have, you'll divide that number by Avogadro's Constant (6.022 x 1023) to deduce the amount of mols of Calcium you have, giving:
$${\mathtt{0.516}}{\mathtt{\,\times\,}}{\mathtt{molCa}}$$
Checking the periodic table shows us that Calcium has a molar weight of 40.08g/mol Ca. Multiply the number of mols, and it's molar weight, giving:
$${\mathtt{0.516}}{\mathtt{\,\times\,}}{\mathtt{molCa}}{\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{40.08}}{\mathtt{\,\times\,}}{\mathtt{gCa}}\right)}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molCa}}\right)}}\right) = {\mathtt{20.68}}{\mathtt{\,\times\,}}{\mathtt{gCa}}$$
Good luck.
