Problem 3 A 34.55 g sample of a pure compound is found to contain 7.12 g of elem

Question

Problem 3 A 34.55 g sample of a pure compound is found to contain 7.12 g of element X and 27.43 g of element Y. If a second sample of the pure compound is found to contain 2.173 of element X, whet is the mass of elernent Y in te seccnd sampler LSubrnit Answer: Irino/5 Problem 4 A sample of a pure compound is found to contain 19.77 g of clement X, 31.17 g of element Y, and 7.78 g of clement Z. If a scoond sample of the pure compound is found to contain 1.88 of clement X, what is the total mass of the sample in the second sample? Submit Answer Trics 0/5 Problem 5 Consider he following isotope 703-2 Hoe many protons, tpt neutrons, and electran s are ttere In this nrepe? tte Submit Answer Trles 05

Explanation / Answer

3) 34.55 g of sample contains 7.12 g of X and 27.43 g of Y

Thus, % composition of X = (7.12/34.55)*100 = 20.61%

The remaing will be composition of Y = 100-20.16 = 79.84%

Now, in the second sample, X is 2.173 g

Mass of second sample = (2.173/(20.61/100)) = 10.543 g

Now, composition of Y in 10.543 g = 10.543 - 2.173 = 8.370 g of Y

Thus the mass of Y in second sample = 8.370 g.

4) Given, Total mass of sample 1 = 19.77+31.17+7.78 = 58.72 g

composition of X = 19.77 g = (19.77/58.72)*100 = 33.67%

composition of Y = 31.17 g = (31.17/58.72)*100 = 53.08%

composition of Z = 7.78 g = 100 - (33.67+53.08) = 13.25%

In second sample, X is 1.88g, then we can use rational:

if 33.67 g of X in 100 g of sample, then 1.88 g of X in x g of sample

x = (1.88*100)/33.67 = 5.58 g of sample

Thus mass of seond sample is 5.58 g .

5) 17O-2 isotope: 17 is the mass number, -2 is the charge and O is oxygen so an atomic number is 8.

Number od protons = atomic number = 8

# of protons + # of Neutrons = mass number = 17

# of Neutrons = 17-8 = 9

# of protons + # of electrons = 0

But we have -2 charge on the atom, hence there are 2 extra electrons

So total # of electrons = 8+2 = 10

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