A toothpaste contains 0.254 % by mass sodium fluoride used to prevent dental caries and 4.4 % by mass KNO3, which decreases pain sensitivity. One tube contains 119 g of toothpaste.

Part A
How many moles of NaF are in the tube of toothpaste?
Express your answer using three significant figures.
Part B
How many fluoride ions, F−, are in the tube of toothpaste?
Express your answer using three significant figures.
Part C
How many grams of sodium ion, Na+, are in 1.35 g of toothpaste?
Express your answer using three significant figures.
Part D
How many KNO3 formula units are in the tube of toothpaste?
Express your answer using two significant figures.
Part E
What is the molar mass of ammonium sulfate?
Express your answer using four significant figures.

Part F

Chloroform, CHCl3, was formerly used as an anesthetic but its use was discontinued due to respiratory and cardiac failure.

How many grams are in 0.133 moles of chloroform?

How many moles are in 35.0 g of chloroform?

A. 7,20x10⁻³mol NaF

B. 4,33x10²¹ F⁻ ions

C. 1,88x10⁻³g Na⁺

D. 3,1x10²² formula units

E. 132,08g/mol

F. 15,9g of CHCl₃; 0,293 mol of CHCl₃

Explanation:

A. 119g × (0,254% / 100) = 0,302g of NaF × (1mol / 41,99g) = 7,20x10⁻³mol NaF

B. 7,20x10⁻³mol NaF ≡ 7,20x10⁻³mol F⁻ × (6,022x10²³ ions / mol) = 4,33x10²¹ F⁻ ions

C. 1,35g × (0,254% / 100) = 0,00343g of NaF × (22,99g Na⁺ / 41,99g NaF) = 1,88x10⁻³g Na⁺

D. 119g × (4,4% / 100) = 5,236g of KNO₃ × (1mol / 101,1g) = 0,0518mol × (6,022x10²³ F.U / 1mol) = 3,1x10²² formula units

E. Formula of ammonium sulfate is (NH₄)₂SO₄. Molar mass is obtained thus:

N: 2N×(14g/mol) = 28g/mol

H: 8H×(1,01g/mol) = 8,08g/mol

S: 1S×(32g/mol) = 32g/mol

O: 4O×(16g/mol) = 64g/mol

Molar mass: 28 + 8,08 + 32 + 64 = 132,08g/mol

F. Molar mass of chloroform is 119,38g/mol, thus:

0,133 mol × (119,38g/mol) = 15,9g of CHCl₃

35,0g  × (1mol / 119,38g) = 0,293mol of CHCl₃

I hope it helps!

can you list the answers so i can answer

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole