Question

question 4
1 pts
calculate the percentage by mass of nitrogen in ptcl2(nh3)2.
12.67
9.34
4.95
9.90
4.67
question 5
1 pts
a 30.5 gram sample of glucose (c6h12o6) contains______mol of glucose.
0.169
2.36
5.90
0.424
0.136

Explanation:

Question 4

Step 1: Calculate the molar mass of \(PtCl_{2}(NH_{3})_{2}\)

• Molar mass of \(Pt = 195.08\space g/mol\)
• Molar mass of \(Cl = 35.45\space g/mol\), so for \(2Cl\): \(2\times35.45=70.9\space g/mol\)
• Molar mass of \(N = 14.01\space g/mol\), so for \(2N\): \(2\times14.01 = 28.02\space g/mol\)
• Molar mass of \(H = 1.008\space g/mol\), so for \(6H\) (\(2\times3H\)): \(6\times1.008=6.048\space g/mol\)
• Molar mass of \(PtCl_{2}(NH_{3})_{2}=195.08 + 70.9+28.02 + 6.048=300.048\space g/mol\)

Step 2: Calculate the mass percentage of \(N\)

• Mass of \(N\) in \(PtCl_{2}(NH_{3})_{2}\) is \(28.02\space g/mol\)
• Mass percentage of \(N=\frac{28.02}{300.048}\times 100\%\approx9.34\%\)

Question 5

Step 1: Calculate the molar mass of \(C_{6}H_{12}O_{6}\)

• Molar mass of \(C = 12.01\space g/mol\), so for \(6C\): \(6\times12.01 = 72.06\space g/mol\)
• Molar mass of \(H = 1.008\space g/mol\), so for \(12H\): \(12\times1.008=12.096\space g/mol\)
• Molar mass of \(O = 16.00\space g/mol\), so for \(6O\): \(6\times16.00 = 96.00\space g/mol\)
• Molar mass of \(C_{6}H_{12}O_{6}=72.06+12.096 + 96.00=180.156\space g/mol\)

Step 2: Calculate the number of moles of \(C_{6}H_{12}O_{6}\)

• Using the formula \(n=\frac{m}{M}\), where \(m = 30.5\space g\) and \(M = 180.156\space g/mol\)
• \(n=\frac{30.5}{180.156}\approx0.169\space mol\)

Answer:

Question 4: 9.34
Question 5: 0.169