One cup of fresh orange juice contains 118 mg of ascorbic acid (vitamin c, c6h8o6). given that one cup = 225.2 ml calculate the molarity of vitamin c in organic juice.

Answer;

The molarity of vitamin C will be; 0.002977 M or 2.977× 10^-3 M

Solution;

0.2252 L = 1 cup

118 mg of ascorbic acid = 0.118 g

1 mole of ascorbic acid = 176 g

Thus; 0.118g C6H8O6 x (1 mol C6H8O6 / 176.0g C6H8O6) / 0.2252L = 2.977x10^-3M .

=0.002977 M or 2.977× 10^-3 M

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally

the correct answer would be b. amelia, because the particals of gas move really fast (charged particals) and gas also has a lot of room (particals move independently) so the only answer that has both of those is amelia's.