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garik1379 [7]
3 years ago
8

If an element has an atomic number of 8 and an atomic mass number of 18, would it be an isotope? Why or why not?

Chemistry
1 answer:
hoa [83]3 years ago
3 0

Answer:

It would be an isotope.

Background Information:

Isotopes are typically elements that have a different number of protons than neutrons. The atomic mass is the total number of protons and neutrons. The atomic number is the number of protons.

Explanation:

If the atomic number is the number of protons that means that this particular element has 8 protons. If the atomic mass is the total number of protons and neutrons then we can simply take away the amount of protons from that number, 18 - 8 = 10. If we take protons away from the number of protons and neutrons we are left with the number of neutrons. So there are 10 neutrons. Because there are 8 protons and 10 neutrons, or a different amount of neutrons and protons we know that this particular atom is an isotope.

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Explanation:

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The atomic mass of carbon -13 is 13. It has six protons. How many neutrons does the isotope have​
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4. In the following, how many digits should be in the solution to have the proper number of sig figs?
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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
Diano4ka-milaya [45]

Answer:

\large \boxed{109.17 \, ^{\circ}\text{C}}

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

2. Kilograms of water

m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}

3. Molal concentration of EG

b =  \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}

4. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}

5. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}

7 0
3 years ago
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