Answer:
Given that camphor is a sublime material while sand is not, a gradual heating of the mixture allows for the separation of camphor from sand. The camphor vapors can then be collected and allowed to cool. This process will result in the formation of solid camphor crystals.
Explanation:
For 1.000 g of X, the mass of Y is 0.1621 g. The mass ratio of Y = 2.100 g: 0.1621 g equals 1:0.07. For 1.000 g of X, the mass of Y is 0.7391 g, which leads to a mass ratio of Y = 2.100 g: 0.7391 g simplifying to 1:0.35 or 20:7. For 1.000 g of X, the corresponding mass of Y is 0.2579 g, yielding a mass ratio of Y = 2.100 g: 0.2579 g resulting in 1:0.12. For 1.000 g of X, the mass of Y becomes 0.2376 g, giving a mass ratio of Y = 2.100 g: 0.2376 g, simplifying to 1:0.11. Lastly, for 1.000 g of X, the mass of Y is determined to be 0.2733 g, leading to a mass ratio of Y = 2.100 g: 0.2733 g which reduces to 1:0.13. Among the calculated values, option B aligns most closely with the law of multiple proportions.
Answer:
The accurate answer is 596.5 kJ.
Explanation:
The question specifies that the mass of ethanol, C2H5OH, is 20 grams.
The molar mass of ethanol is 46 g/mol.
To find the moles of ethanol, we use the formula:
n = mass / molar mass
= 20/46 = 0.435 moles
According to the question, the standard heat of combustion for ethanol is 1372 kJ/mol. Hence, one mole releases 1372 kilojoules during combustion.
The energy produced from burning 20 grams of ethanol completely is 0.435 * 1372 = 596.5 kJ.
<span>Using PV=nRT, which represents a universal constant for any state, we have:
P1V1/n1T1=R
and
P2V2/n2T2=R;
This implies that:
P1V1/n1T1=P2V2/n2T2
Thus we can express it as
V1/n1=V2/n2.
Rearranging yields:
V2=V1 x (n2/n1) = 750 mL x ((0.65+0.35)/(0.65)) = 1200 mL = 1.2 L... with 2 significant figures</span>