Answer:
1.71 × 10²¹ molecules
Explanation:
1 gram is equal to 1000 milligrams. The mass in grams corresponding to 500 mg is:
500 mg × (1 g / 1000 mg) = 0.500 g
The molar mass of ascorbic acid is 176.12 g/mol. The moles corresponding to 0.500 grams of ascorbic acid are:
0.500 g × (1 mol/ 176.12 g) = 0.00284 mol
In 1 mole of ascorbic acid, there are 6.02 × 10²³ molecules of ascorbic acid (Avogadro's number). The molecules in 0.00284 moles are:
0.00284 mol × (6.02 × 10²³ molecules/ 1 mol) = 1.71 × 10²¹ molecule
Answer : The molal freezing point depression constant of X is 
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of X liquid (solvent) = 450.0 g
Molar mass of urea = 60 g/mole
Formula used :
where,
= change in freezing point
= freezing point of solution = 
= freezing point of liquid X= 
i = Van't Hoff factor = 1 (for non-electrolyte)
= molal freezing point depression constant of X = ?
m = molality
Now put all the given values in this formula, we get
![[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}](https://tex.z-dn.net/?f=%5B0.4-%28-0.5%29%5D%5EoC%3D1%5Ctimes%20k_f%5Ctimes%20%5Cfrac%7B5.90g%5Ctimes%201000%7D%7B60g%2Fmol%5Ctimes%20450.0g%7D)
Therefore, the molal freezing point depression constant of X is
I got 0 ?? if that’s any help
Answer:
Decay-the breakdown of dead plants..
Earth- thermal energy comes from deep inside...
Fires- these consume feul...
Explanation:
So in order for us to know the percentage of sugar present in a 12.00 g of milk chocolate, what we are going to do is that, we just have to divide 8 by 12 and multiply in by 100 and we get 66.67. Therefore, the percentage of sugar present in 12.00 g of milk chocolate bar is 66.67%. Hope this answers your question. Have a great day!
