Question

multiple choice
identify the choice that best completes the statement or answers the question.

1. calculate the number of molecules in 0.0047 g of lithium carbonate, li₂co₃.

a. 2.1 x 10²⁴ molecules li₂co₃
b. 3.8 x 10²⁰ molecules li₂co₃
c. 2.1 x 10²³ molecules li₂co₃
d. 3.8 x 10¹⁹ molecules li₂co₃

use this table to answer the next four questions.

extended symbol	atomic number	mass number	protons	neutrons	electrons	charge
ca		40		(ii)		0
(iii)			9	10		-1
			13	14	11	(iv)

1. the extended symbol roman numeral iii is…

a. $\ce{^{18}_{9}f}$
b. $\ce{^{9}_{19}f^{-1}}$
c. $\ce{^{19}_{9}f^{-1}}$
d. $\ce{^{10}_{9}f^{-1}}$
e. $\ce{^{9}_{9}f^{-1}}$

1. the charge for roman numeral iv is…

a. +13
b. 0
c. +1
d. -2
e. +2

1. the number of electrons for roman numeral i is…

a. 3
b. 11
c. 27
d. 13
e. 14

1. the number of neutrons roman numeral ii is…

a. 0
b. 2
c. 20
d. 40
e. 60

1. an aluminum isotope consists of 13 protons, 13 electrons, and 14 neutrons. its mass number is…

a. 13
b. 14
c. 27
d. 40

Explanation:

Step1: Calculate molar mass of $\text{Li}_2\text{CO}_3$

Molar mass $= 2\times6.94 + 12.01 + 3\times16.00 = 73.89 \ \text{g/mol}$

Step2: Find moles of sample

$n=\frac{0.0047}{73.89} \approx 6.36\times10^{-5} \ \text{mol}$

Step3: Calculate number of molecules

$\text{Molecules} = 6.36\times10^{-5} \times 6.022\times10^{23} \approx 3.8\times10^{19}$
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Step4: Solve Q2: Identify species III

Mass number $=9+10=19$, atomic number $=9$, charge $=-1$. Symbol is $_{9}^{19}\text{F}^{-1}$
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Step5: Solve Q3: Find charge IV

Charge $=13-11=+2$
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Step6: Solve Q4: Find electrons I

Electrons $=13-2=11$
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Step7: Solve Q5: Find neutrons II

Neutrons $=40-20=20$ (Ca has 20 protons)
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Step8: Solve Q6: Find mass number

Mass number $=13+14=27$

Answer:

1. d. $3.8 \times 10^{19}$ molecules $\text{Li}_2\text{CO}_3$
2. c. $_{9}^{19}\text{F}^{-1}$
3. e. $+2$
4. b. $11$
5. c. $20$
6. c. $27$