CHEM

2 answers:

3 0

Molarity = moles / liter

a) M = 2/4 = 0.5 M

b) Moles = 4/(30 + 16 + 1)
= 0.085
M = 0.085 / 2 = 0.0425 M

c) Moles = 5.85 / (23 + 35.5)
= 0.1
M = 0.1 / 0.4
= 0.25 M

Dmitrij [34]3 years ago

3 0

Molarity =moles of solute /volume of solution. A.) Molarity of glucose =moles of glucose /volume of solution =2/4 = 0.5 M B.) No of moles of KOH= weight of KOH / molecular weight of KOH = 4/ 56.10 =0.07 moles. Therefore, Molarity of KOH =moles of KOH /volume of solution = 0.07/2 = 0.035M. C.) No of moles of NaCl = weight of NaCl/ molecular weight of NaCl = 5.85/58.5 = 0.1moles. Therefore, Molarity of NaCl = moles of NaCl /volume of solution = 0.1/ 400 x 10^-3 = 0.25M

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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:

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