Chemistry, 16.07.2021 06:30, truesarah111

Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant figures in calculations. Please show work! Convert 192 grams of phosphorus pentabromide to molecules.

Convert 3.42 kilograms of table sugar (C₁₂H₂₂O₁₁) to molecules.

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answered: genyjoannerubiera

1) 2.69 * 10²³ PBr₅

2) 6.02 * 10²⁴ C₁₂H₂₂O₁₁

Explanation:

Question 1)

We want to convert 192 grams of phosphorus pentabromide to molecules. Note that 192 is three significant figures.

Phosphorus pentabromide is given by PBr₅.

To convert from grams to molecules, we can convert from grams to moles first, and then from moles to molecules.

To convert from grams to moles, we will find the molar mass of PBr₅.

Since the molar mass of P is 30.974 g/mol and the molar mass of Br is 79.904 g/mol, the molar mass of PBr₅ is:

$(30.974)+5(79.904) = 430.494\text{ g/mol}$

And since we want to convert from grams to moles, we can write the following ratio:

$\displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}$

Where grams is in the denominator, which allows us to cancel them out, leaving us with only moles.

To convert from moles to molecules, we can use the definition of the mole: a mole of one substance has 6.022 * 10²³ amount of that substance.

So, a mole of PBr₅ has 6.022 * 10²³ molecules of PBr₅. Since we want to cancel out the moles, we can write the ratio:

$\displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}$

In combination, starting with 192 grams of PBr₅, we will acquire:

$\displaystyle 192\text{ g PBr$_5$} \cdot \displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}$

Cancel like units:

$\displaystyle = 192 \cdot \displaystyle \frac{1 }{430.494}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1}$

Multiply. Hence:

$=2.6858...\times 10^{23}\text{ PBr$_5$}$

Since the final answer should have three significant digits, our final answer is:

$= 2.69\times 10^{23} \text{ PBr$_5$}$

So, there are about 2.69 * 10²³ molecules of PBr₅ in 192 grams of the substance.

Question 2)

We want to convert 3.42 kilograms of table sugar (C₁₂H₂₂O₁₁) to molecules. Note that this is three significant figures.

3.42 kilograms is equivalent to 3420 grams of table sugar.

Again, we can convert from grams to moles, and then from moles to molecules.

First, we will find the molar mass of table sugar. The molar mass of carbon is 12.011 g/mol, hydrogen 1.008 g/mol, and oxygen 15.999 g/mol. Thus, the molar mass of table sugar will be:

$12(12.011)+22(1.008)+11(15.999) = 342.297\text{ g/mol}$

To cancel units, we can write our ratio as:

$\displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}$

With grams in the denominator.

And by definition:

$\displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}$

Combining the two ratios and the starting value, we acquire:

$3420 \text{ g C$_{12}$H$_{22}$O$_{11}$}\cdot \displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}$

Cancel like units:

$=3420 \cdot \displaystyle \frac{1}{342.297}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1}$

Multiply:

$\displaystyle = 60.1677... \times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}$

Rewrite:

$\displaystyle = 6.01677... \times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}$

The resulting answer should have three significant digits. Hence:

$=6.02\times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}}$

So, there are about 6.02 * 10²⁴ molecules of table sugar in 3.42 kilograms of the substance.

answered: bvaughn5915

2.69×10²³ molecules of PBr₅

6.02×10²⁴ molecules of C₁₂H₂₂O₁₁

Explanation:

To solve the first problem, we want to first find formula for phosphorus pentabromide, which is PBr₅. Now, we need to know the molar mass of PBr₅, which is about 430.49 g/mol. To get to molecules, we need to use Avogadro's number, which is 6.022×10²³ molecules/mol.

$192g*\frac{1mol}{430.49g} *\frac{6.022*10^{23}molecules}{1mol} =2.69*10^{23} molecules$

Now, we know that there are about 2.69×10²³ molecules of PBr₅.

To solve the second problem, we need to use Avogadro's number, along with finding the molar mass of C₁₂H₂₂O₁₁, and converting kilograms to grams.

$3.42 kg*\frac{1000g}{1kg} *\frac{1mol}{342.3g} *\frac{6.022*10^{23} molecules}{1mol} =6.02*10^{24} molecules$

Now, we know that there are about 6.02×10²⁴ molecules of C₁₂H₂₂O₁₁.

answered: Guest

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation:

answered: Guest

Ch4 and ph3 are both covalent since they are composed both of non-metals. ch4 is non- polar because of london forces and has a geometry of tetrahedral ph3 , on the other hand, is umbrella shaped polar and has dipole-dipole intermolecular force (imf) . since ph3 is polar, it has a stronger imf