3. What mass of the compond CrO 3 (M=100.0) conatins 4.5 x 10 23 Oxygen atoms? 4

Question

3. What mass of the compond CrO3 (M=100.0) conatins 4.5 x 1023 Oxygen atoms?

4. An 18.5 g sample of tin (M=118.7) combines with 10.0 g of sulfur ( M=32.07) to form a compound. what is the empirical formula?

5. A mixture is prepared by adding 50.0 mL of 0.200 M NaOH to 75.0 mL of 0.100 M NaOH. What is the [OH-] in the mixture?

6. What mass of NaHCO3 ( M=84.0) is required to completely neutralize 25.0 mL of 0.125 M H2SO4?

9. Which cation forms an insoluble chloride and insoluble sulfide?

(a) Ba2+ (b) Cu2+ (c) Mn2+ (d) Pb2+

10. Which 0.10 M (aq) soln. exhibits the lowest electrical conductivity?

(a) HC2H3O2 (b) HNO3

15. These substances have the same general molar mass, which has the lowest boiling point?

(a) CH3CH2CH3   (b) CH3OCH3 (c) CH3CH2OH

PLEASE EXPLAIN EVERYTHING FOR FULL CREDIT!

Explanation / Answer

3) moles of oxygen = no of atoms / avagadro number

moles of oxygen = 4.5 x 1023 / 6.023 x 1023

moles of oxygen atoms = 0.747

1 mole of Cr03 contains 3 moles of oxygen atoms

so

moles of Cr03 = moles of oxygen atoms / 3

moles of Cr03 = 0.747 /3

moles of Cr03 = 0.249

mass = moles x molar mass

mass of Cr03 = 0.249 x 100

mass of Cr03 = 24.9 grams

4)

moles of tin = mass / molar mass

moles of tin = 18.5 / 118.7

moles of tin = 0.1558

moles of sulfur = 10/32.07

moles of sulfur = 0.3118

ratio of tin / sulfur = 0.1558/0.3118

ratio of tin moles / sulfur moles = 1/2

so for 1 mole of tin there are 2 moles of sulfur

so the empirical formula is SnS2

5) moles of NaOH 1 = molarity x volume /1000

moles of NaOH 1 = 0.2 x 50 /1000 = 0.01

moles of NaOH 2 = 0.1 x 75 /1000 = 0.000.0175 75

total moles = 0.01 + 0.075

total moles = 0.0175

total volume = 50 + 75 = 125

conc of NaOH = molesx 1000/ volume

conc of NaoH = 0.0175 x 1000/ 125

conc of NaOH = 0.14 M

as NaOH is a strong base

[OH-] = [NaOH} = 0.14 M

so

[OH-] = 0.14 M

6)     2NaHC03 + H2S04 ----> Na2S04 + 2H2C03

so

moles of NaHC03 = 2 x mole sof H2S04

moles of NaHC03 = 2 x 0.125 x 25 /1000

moles of NaHC03 = 6.25 x 10-3

mass of NaHC03 = moles x molar mass

mass of NaHC03 = 6.25 x 10-3 x 84

mass of NaHC03 = 0.525 g

9) d) Pb+2

10) a ) HC2H302

15)   (a) CH3CH2CH3

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