It is not a pure substance, because a solution are mixed chemicals in a way that the molecules are not bonded with one another. Thus, separating them from compounds and elements, which are pure substances.

8 0

3 years ago

What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o

lyudmila [28]

Answer:

$\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%$

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

$atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu$

Thus, the atom percent turns out:

$\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%$

Best regards.

4 0

3 years ago

A chemist prepares a solution of mercury(I) chloride Hg2Cl2 by measuring out

Genrish500 [490]

Answer:

1.26 × 10^-8 M

Explanation:

We are given;

Number of moles of mercury (i) chloride as 0.000126 μmol

Volume is 100 mL

We are required to calculate the concentration of the solution.

We need to know that;

Concentration is also known as molarity is given by;

Molarity = Number of moles ÷ Volume

Number of moles = 1.26 × 10^-10 Moles

Volume = 0.01 L

Therefore;

Concentration = 1.26 × 10^-10 Moles ÷ 0.01 L

= 1.26 × 10^-8 M

Thus, the molarity of the solution is 1.26 × 10^-8 M

6 0

3 years ago