Monomers combine through electron sharing during the polymerization process. This leads to the formation of a polymer, which consists of repeating units. The resulting substance has various applications.
The result is 200 g. Given that the molar mass of CaCl2 is 110.98 g/mol, this indicates that there are 110.98 g in 1 L of a 1 M solution. Let's calculate the amount of CaCl2 in 0.720 M. Using the proportion 110.98 g: 1 M = x: 0.720 M, we find x to be 79.90 g. Therefore, in 1 L of a 0.720 M solution, there is 79.90 g. Next, we need to create ten beakers with 250 mL each, totaling 10 * 250 mL = 2500 mL or 2.5 L. Then, using the equation 79.90 g: 1 L = x: 2.5 L, we calculate x = 79.90 g * 2.5 L: 1 L, resulting in x = 199.75 g, approximately 200 g.
<span>(NH4)2CO3 -> 96.09 g/mol
(6.995g ammonium carbonate)(1mol ammonium carbonate/ 96.09 g ammonium carbonate) = 0.072796 mol ammonium carbonate
In this calculation, the unit 'grams' cancels out as it's present in both the numerator and the denominator, leading to 'mol' being the remaining unit.
Examining the formula (NH4)2CO3, it can be interpreted as:
2 mol (NH4) + 1 mol (CO3) = 1 mol (NH4)2CO3
This means every mole of ammonium carbonate yields one mole of carbonate ions and two moles of ammonium ions.
(0.072796 mol ammonium carbonate) = (0.072796 mol carbonate ion) + (0.363981 mol ammonium ion) </span>
