In 1 mole of
, there are 3 moles of ions, 1 mole of Ca^{2+} and 2 mole of
.

Molar mass of
is 110.98 g/mol. Calculating number of moles from given mass as follows:

Thus, number of moles of ions will be
.
Since, 1 mole of any substance has
units of that substance where
is Avogadro's number.
Thus, 9.38 mol of ions will have
number of ions.
Therefore, total number of ions in 347 g of
is
.
Answer:
1.66 * 10^-18 mol of Argon
Explanation:
Avogadro says 1 mol of anything has 6.022* 10^23 atoms/particles. Thus using this principle, 1 million argon atoms will have


Answer: -
Surface Tension
Explanation: -
Surface tension is cohesive force created as a result of hydrogen bonding, that enables a liquid drop to have a minimum surface area.
Due to it being cohesive, the water top surface is concave in nature, allowing us to hence slightly overfill a glass with water.
Due to surface tension, the surface of water behaves like a stretched membrane, allowing dense objects like a length wise steel needle to float on water.
Thus, the hydrogen bonding in water creates __surface tension__, a cohesive force that enables one to slightly overfill a glass with water or allows denser objects, such as a lengthwise steel needle, to float on water
<span>0.50 mg x ( 1 mL / 0.25 mg ) = 2 mL </span>
<span>If there are 38 g of CFC, then there will be 120.9135 g of CFCl2 per mol. You will multiply this times the number of moles of Cl (2) for every mole of CFCl2, and then by the number of grams of Cl per mole, which is 35.4532: (38.0 g CF2Cl2) / (120.9135 g CF2Cl2/mol) x (2 mol Cl / 1 mol CF2Cl2) x (35.4532 g Cl/mol) =
22.3 g Cl in CF2Cl2
Next, if there are 38 g of CFC, there will be 137.3861 g of CFCl3 per mole. You will multiply this times the number of moles of Cl (3 this time) for every mole of CFCl3. You will then multiply this by 35.4532 again:
(38.0 g CFCl3) / (137.3681 g CFCl3/mol) x (3 mol Cl / 1 mol CFCl3) x (35.4532 g Cl/mol) =
29.4 g Cl in CFCl3
Continue following these steps until you are able to multiply 1 mole of Cl per 1 mol CF3Cl by 35.4532:
(38.0 g C2F3Cl3) / (187.3756 g C2F3Cl3/mol) x (3 mol Cl / 1 mol C2F3Cl3) x (35.4532 g Cl/mol) =
21.6 g Cl in C2F3Cl3
(38.0 g CF3Cl) / (104.4589 g CF3Cl/mol) x (1 mol Cl / 1 mol CF3Cl) x (35.4532 g Cl/mol) =
12.9 g Cl in CF3Cl</span>