The molecular mass of Carvone is calculated as;
= 12 (C)₁₀ + 1.008 (H)₁₄ + 16 (O)
= 120 + 14.112 + 16
= 150.112
%age of Carbon;
= (120 ÷ 150.112) × 100
= 79.94 %
%age of Hydrogen;
= (14.112 ÷ 150.112) × 100
= 9.40 %
%age of Oxygen;
= (16 ÷ 150.112) × 100
= 10.65 %
Answer:
D
Explanation:
If the pressure remains constant then the temperature and Volume are all that you have to consider.
Givens
T1 = 19oC = 19 + 273 = 292o K
T2 = 60oC = 60 + 273 = 333oK
V1 = 250 mL
V2 = x
Formula
V1/T1 = V2/T2
250/292 = x/333
Solution.
The solves rather neatly. Multiplly both sides by 333
250*333 / 292 = 333 *x / 333
Do the multiplication
250 * 333 / 292 = x
83250 / 292 = x
Divide by 292
x = 285.1 mL
The answer is D
Answer:
First of all, it's KNO₃ not KNO.
Second, KNO₃ is neither an acid nor it is a base, infact, it is a salt and therefore it's neutral.
hope that helps...
<span>Divide the number of grams present in the sample by copper's gram atomic weight to find the number of gram atomic weights present. Then multiply that result by Avogadro's Number: 6.022137 x 10^23 atoms/gram atomic weight.1,200 g/(63.54 g/gram atomic weight) ? 18.885741 gram-atomic weights. Hope this helps. </span>
