Chemistry

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GED Science › Chemistry

Questions 1 - 10

How many molecules are in 3.72mol of fluorine?

$2.24\times10^{24}\ \text{molecules}$

$6.18\times10^{-24}\ \text{molecules}$

$70.68\ \text{molecules}$

$0.20\ \text{molecules}$

Explanation

Convert from moles to molecules using Avogadro's number:

$(\text{moles})\times(\frac{6.02\times10^{23}\text{molecules}}{\text{mole}})=\text{molecules}$

Use the moles given in the question to solve.

$3.72mol\times \frac{6.02\times10^{23}\ \text{molecules}}{1mol}=2.24\times10^{24}\ \text{molecules}$

How many molecules are in 3.72mol of fluorine?

$2.24\times10^{24}\ \text{molecules}$

$6.18\times10^{-24}\ \text{molecules}$

$70.68\ \text{molecules}$

$0.20\ \text{molecules}$

Explanation

Convert from moles to molecules using Avogadro's number:

$(\text{moles})\times(\frac{6.02\times10^{23}\text{molecules}}{\text{mole}})=\text{molecules}$

Use the moles given in the question to solve.

$3.72mol\times \frac{6.02\times10^{23}\ \text{molecules}}{1mol}=2.24\times10^{24}\ \text{molecules}$

How many molecules are in 3.72mol of fluorine?

$2.24\times10^{24}\ \text{molecules}$

$6.18\times10^{-24}\ \text{molecules}$

$70.68\ \text{molecules}$

$0.20\ \text{molecules}$

Explanation

Convert from moles to molecules using Avogadro's number:

$(\text{moles})\times(\frac{6.02\times10^{23}\text{molecules}}{\text{mole}})=\text{molecules}$

Use the moles given in the question to solve.

$3.72mol\times \frac{6.02\times10^{23}\ \text{molecules}}{1mol}=2.24\times10^{24}\ \text{molecules}$

A student is asked to make $\small 4L$ of $\small 0.125M$ NaCl solution. The student is given a stock solution of $\small 5M$ NaCl. What volumes of stock NaCl solution and pure water could be used to make the desired final solution?

$100 mL\ \text{stock NaCl and}\ 3900mL\ \text{water}$

$100 mL\ \text{stock NaCl and}\ 4000mL\ \text{water}$

$125 mL\ \text{stock NaCl and}\ 3875mL\ \text{water}$

$125 mL\ \text{stock NaCl and}\ 4000mL\ \text{water}$

Explanation

Convert liters to mL:

$\small \frac{4L}{1}=\frac{1000mL}{1L}=4000 mL$

This is the TOTAL volume of the solution. Next, determine how much NaCl to add.

$\small c_1v_1=c_2v_2$

$\small c_1=5.0M$

$\small c_2=0.125M$

$\small v_2=4000mL$

$\small (5.0M)v_1=(0.125M)(4000mL)$

$\small (5.0M)v_1=500MmL$

$\small \frac{(5.0M)v_1}{5.0M}=\frac{500MmL}{5.0M}$

$\small v_1=100mL$

Therefore, $\small 100mL$ of stock NaCl must be added. Subtract this volume from $\small 4000 mL$ to determine the amount of water used in the final solution.

$\small 4000mL-100mL=3900mL\ water$

$100 mL\ \text{stock NaCl and}\ 3900mL\ \text{water}$

$100 mL\ \text{stock NaCl and}\ 4000mL\ \text{water}$

$125 mL\ \text{stock NaCl and}\ 3875mL\ \text{water}$

$125 mL\ \text{stock NaCl and}\ 4000mL\ \text{water}$

Explanation

Convert liters to mL:

$\small \frac{4L}{1}=\frac{1000mL}{1L}=4000 mL$

This is the TOTAL volume of the solution. Next, determine how much NaCl to add.

$\small c_1v_1=c_2v_2$

$\small c_1=5.0M$

$\small c_2=0.125M$

$\small v_2=4000mL$

$\small (5.0M)v_1=(0.125M)(4000mL)$

$\small (5.0M)v_1=500MmL$

$\small \frac{(5.0M)v_1}{5.0M}=\frac{500MmL}{5.0M}$

$\small v_1=100mL$

Therefore, $\small 100mL$ of stock NaCl must be added. Subtract this volume from $\small 4000 mL$ to determine the amount of water used in the final solution.

$\small 4000mL-100mL=3900mL\ water$

$100 mL\ \text{stock NaCl and}\ 3900mL\ \text{water}$

$100 mL\ \text{stock NaCl and}\ 4000mL\ \text{water}$

$125 mL\ \text{stock NaCl and}\ 3875mL\ \text{water}$

$125 mL\ \text{stock NaCl and}\ 4000mL\ \text{water}$

Explanation

Convert liters to mL:

$\small \frac{4L}{1}=\frac{1000mL}{1L}=4000 mL$

This is the TOTAL volume of the solution. Next, determine how much NaCl to add.

$\small c_1v_1=c_2v_2$

$\small c_1=5.0M$

$\small c_2=0.125M$

$\small v_2=4000mL$

$\small (5.0M)v_1=(0.125M)(4000mL)$

$\small (5.0M)v_1=500MmL$

$\small \frac{(5.0M)v_1}{5.0M}=\frac{500MmL}{5.0M}$

$\small v_1=100mL$

Therefore, $\small 100mL$ of stock NaCl must be added. Subtract this volume from $\small 4000 mL$ to determine the amount of water used in the final solution.

$\small 4000mL-100mL=3900mL\ water$

On the periodic table, carbon is atomic number $\small 6$ and has an atomic mass of $\small 12.01amu$ . How many sub-atomic particles are found in the nucleus of a carbon atom?

$\small 12$

$\small 6$

$\small 18$

$\small 2$

Explanation

The standard sub-atomic particles are protons, neutrons, and electrons. Protons and neutrons are found in the nucleus, while electrons move freely around the nucleus. This question is specifically asking about the number of protons and neutrons in an atom of carbon.

Each proton and each neutron has a mass of approximately . Since the atomic mass of a carbon atom is $\small 12.01amu$ , we can conclude that there are twelve sub-atomic particles in the nucleus (twelve total protons and neutrons).

The atomic number refers to the number of protons in the nucleus of the atom. Since the atomic number of carbon is six, there are six protons in the nucleus. For this question, however, we need the total of both the protons and neutrons (not just the protons).

On the periodic table, carbon is atomic number $\small 6$ and has an atomic mass of $\small 12.01amu$ . How many sub-atomic particles are found in the nucleus of a carbon atom?

$\small 12$

$\small 6$

$\small 18$

$\small 2$

Explanation

On the periodic table, carbon is atomic number $\small 6$ and has an atomic mass of $\small 12.01amu$ . How many sub-atomic particles are found in the nucleus of a carbon atom?

$\small 12$

$\small 6$

$\small 18$

$\small 2$