Question

what is the value of the following operation to the correct sig figs and units? do not include the chemical formula in the units as you would normally do.
8.346 g pcl₃ (1 mol pcl₃ / 137.33 g pcl₃) (3 mol cl₂ / 6 mol pcl₃) (70.906 g cl₂ / 1 mol cl₂) = units
question 9
1 pts
solve the following problem using the correct significant digits and units.
(0.02704 mol)(0.082058 atm l k⁻¹ mol⁻¹)((15.7 + 273.15)k) / 0.1544 l = units
question 10
1 pts
solve the following problem without the use of a calculator to the correct significant digits and units. note that you should not need a calculator to solve this problem.
2.6×10⁻¹⁹ mm×(10⁻³ m / 1 mm)(1 nm / 10⁻⁹ m) = units
to report your answer in scientific notation, follow the following examples:
report 1.23×10⁻³⁴ as 1.23e - 34 or report 1.23×10¹² as 1.23e12
for units raised to a power, use the carrot, ^, to indicate. for example, m³ would be given as m^3

Explanation:

Question 8

Step1: Calculate moles of $PCl_3$

$n_{PCl_3}=\frac{8.346\ g}{137.33\ g/mol}=0.06077\ mol$

Step2: Calculate moles of $Cl_2$ using mole - ratio

The mole - ratio of $PCl_3$ to $Cl_2$ is $6:3 = 2:1$. So $n_{Cl_2}=\frac{3}{6}\times0.06077\ mol = 0.030385\ mol$

Step3: Calculate mass of $Cl_2$

$m_{Cl_2}=n_{Cl_2}\times M_{Cl_2}=0.030385\ mol\times70.906\ g/mol = 2.152\ g$

Step1: Simplify the numerator

First, calculate the value inside the parentheses: $15.7 + 273.15=288.85\ K$
The numerator is $(0.02704\ mol)\times(0.082058\ atm\ L\ K^{-1}\ mol^{-1})\times(288.85\ K)$
$nRT=0.02704\times0.082058\times288.85\ atm\ L=0.6429\ atm\ L$

Step2: Calculate the final result

$\frac{0.6429\ atm\ L}{0.1544\ L}=4.164\ atm$

Step1: Convert units

$2.6\times10^{- 19}\ mm\times(10^{-3}\ m/1\ mm)\times(1\ nm/10^{-9}\ m)$
$2.6\times10^{-19}\times10^{-3}\ m\times\frac{1\ nm}{10^{-9}\ m}$
Using the rule of exponents $a^m\times a^n=a^{m + n}$, we have $2.6\times10^{-19-3}\ m\times\frac{1\ nm}{10^{-9}\ m}=2.6\times10^{-22}\ m\times\frac{1\ nm}{10^{-9}\ m}$
$2.6\times10^{-22 + 9}\ nm=2.6\times10^{-13}\ nm$

Answer:

$2.15\ g$

Question 9