### How many grams are there in 3.58 short tons? Show all work, all units, 1lb= 16 oz, 1 short ton= 2000lb.

3.58 short ton= 2000 lb/ 1 short ton X 16 oz./ 1lb X 1 g/ 0.03527 oz = 3.25E 6 grams

Chap. 1

###
Sample A- Mass of Sample:1.518g MofFE:1.094g MofO: 0.424g

Sample B- 2.056, 1.449g, 0.607g

Sample c- 1.873g, 1.335g, 0.538g

Sample A- 1.094/ 0.424= 2.58

Sample B- 1.449/ 0.607= 2.38

Sample C- 1.335/ 0.538= 2.48

**No because the ratios of M(iron) to M(oxygen) are all different.**

Chap. 1

### The density of osmium metal ( a platinum- group metal) is 22.5 g/cm^3. Express the density in SI units (kg/m^3).

22.5g/cm^3 = 10^-3 kg/1g x 10^6 cm^3/ 1m^3= 2.25x10^4 kg/m^3

Chap. 1

###
Calculate the relative abundance of each isotope of Boron from the following data:

Isotope Mass: Isotope Mass (amu): Average Atomic

B-10: 10.013=

------------------------------ 10.812

B-11: 11.009=

B-10= x

B-11= y

*x+y= 1--> x= 1-y

** x(10.013)+y(11.009)= 10.812

----> (1-y)(10.013)+ 11.009y= 10.812

-----> 10.013- 10.013y+11.009y = 10.812

-10.013 -10.013

----> .996y=.799

***-----> y=.8022088353X100= 80.221% (B-11)

**** B-10= 100%- 80.221%= 19.780%

Chap. 1

###
An element has three naturally occurring isotopes with the following masses and abundances:

Isotopic Mass (amu)----- Fractional Abundance

I- 38.964------ 0.9326

II- 39.964------ 1.000X 10^-4

III- 40.962------ 0.0673

(Isotope Mass)(Fractional Abundance)=

I- 38.964(0.9326) = 36.3378264

II- 39.964(1.000X 10^-4)= .0039964

III- 40.962(0.0673)= 2.7567426

----> I + II + III= 39.0985654

**------> 39.10 Potassium

Chap. 1

### A 1.50 g sample of nitrous oxide (an anesthetic, sometimes called laughing gas) contains 2.05 X 10^22 N2O molecules. How many nitrogen atoms are in this sample? How many nitrogen atoms are in 1.00 g of nitrous oxide?

NO= 1.50g

N2O= 2.05 x 10^22

*---> 1.50g NO/ 2.05X10^22= 1g/ x

------> 1.3666667 X 10^22 nitrogen atoms--> 1.37 X 10^22

= ------> 2(1.37x10^22) = 2.74x 10^22 nitrogen atoms in 1 g of N2O

** Nitrogen=

----> 2.05x10^22 N2O= 2N atoms/1 N2O= 4.10x10^22 nitrogen atoms

Chap. 1

###
How many significant figures are there in the value 4,750,330?

a. 7

b. 6

c. 5

d. 4

e. 3

b. 6

Chap. 1

###
Which of the following sets of numbers have the same number of sig figs?

1. 3.55E-3 2. 3.55E3 3. 3,550,000

a. 1 and 2

b. 1 and 3

c. 2 and 3

d, 1,2,3

e. none of them

d. 1,2,3

Chap. 1

###
The number of significant figures that should be reported for the answer the mathematical computation 142.000- 41.9903 is

a. 2

b. 3

c. 4

d. 5

e. 6

d. 5

Chap. 1

###
Liquid propane boils at -42C. What is its boiling point on the Kelvin scale?

a. 231k

b. 256k

c. 273k

d. 315k

e. 345k

a. 231k

Chap. 1

### The average velocity of oxygen molecule at 1000 is 8X 10^4 cm/s. Which of the following calculations would convert this value to the velocity in miles per hour?

e. 8X10^4 cm/s X 1 in/ 2.54 cm X 1 ft/ 12 in X 1 mi/ 5280 ft X 3600s/ 1 hr

Chap. 1

###
A piece of indium weighing 15.442g is placed in 49.7 cm^3 of ethyl alcohol (d= 0.789 g/ cm^3) in a graduated cylinder. The alcohol level increases to 51.8 cm^3. The best value for density of indium from these data:

a. 7.353 g/ cm^3

b. 7.35 g/cm^3

c. 7.4 g/ cm^3

d. 9.21 g/ cm^3

e. 9.2 g/ cm^3

c. 7.4 g/cm^3

Chap. 1

###
How many protons, neutrons, and electrons are in the tin(II) ion 119, 50 Sn^ 2+? (p,n,e)

a. 119, 50, 119

b. 50, 69, 50

c. 50, 69, 48

d. 69, 50, 69

e. 50, 119, 52

c. 50p, 69n, 48e

Chap. 1

###
A certain element is listed as having 63.5 atomic mass units. It is probably true that it:

a. a mixture of isomers

b. a mixture of allotropes

c. a mixture of neutrons

d. a mixture of ions

e. a mixture of isotopes

e. a mixture of isotopes

Chap. 1

### The mass spectrum of an element with two naturally occurring isotopes is shown below. Its atomic mass would be best estimated as (chart).

c. 63.6

Chap. 1

###
Choose the pair of names and formulas that DO NOT match:

a. Sodium Sulfite Na2S

b. Calcium fluoride CaF2

c. Potassium permanganate KMnO4

d. Aluminum bromide AlBr3

e. Iron (III) oxide Fe2O3

a. Sodium Sulfite--- Na2S

Chap. 1

###
The formula neodymium sulfate is Nd2(SO4). On the basis of this information, the formula for the nitride of neodymium would be expected to be:

a. Nd2(NO2)3

b. Nd2N2

c. Nd(NO3)3

d. Nd(NO2)3

e. NdN

e. NdN

Chap. 1

###
Treatment of sodium borohydrate with sulfuric acid is a convenient method for the preparation of diborane:

__ NaBH4+ ___H2SO4---. ___ B2H6+ H2+ Na2SO4

a. 1

b. 2

c. 3

d. 4

e. 5

b.2

###
Which of the following equations is (are) balanced?

1. NaCl + Pb(NO3)2---> PbCl2 + NaNO3

2. (NH4)2Cr2O7 ---> N2 + 4H2O + Cr2O3

3. FeO4 +3CO ---> 3Fe + 3 Co2

a. 1

b. 2

c. 3

d. 1 and 2

e. 1,2, 3

b. 2

###
What is the ratio of oxygen atoms to hydrogen atoms in the mineral carnotite K2(UO2)3 (VO4)2 X 3H2O:

a. 8:3

b. 8:6

c. 9:6

d. 17:3

e. 17:6

e. 17:6

**END CHAP.1**

###
Calculate the percentage composition for each of the following componds (to three sig digs):

(a) CO2

*C+O2= 12.01 + 2(15.99) = 44.008

** C- 12.01/ 44.008= .273 X 100= 27.3%

*** O2 - 2(15.99)/ 44.008 = .727 X 100= 72.7%

### Malonic acid is used in the manufacture of barbiturates (sleeping pills). The composition of the acis is 34.6% C, 3.9% H, and 61.5% O. What is Malonic acid's empirical formula?

*C- 34.6g x 1mol/ 12.01g C = 2.88

H- 3.9g H X 1mol/ 1.01g H= 3.86

O- 61.5 O x 1 mol/ 15.999g O = 3.84

** C- 2.88/ 2.88= 1 X 3 = 3

H- 3.68/2.88= 1.34 X 3= 4.02--> 4

O- 3.84/ 2.88= 1.33 X 3 = 4

**** C3H4O4 ******

###
Nickel (II) cholride reacts with sodium phosphate to precipitate nickel (II) phosphate and another.

(a) Write a balanced equation for this reaction.

(b) What is the amount (in grams) of nickel (II) phosphate produced from 42.6g of sodium phosphate?

(a) 3NiCl2 + 2Na3PO4 ---> Ni3(PO4)2 + 6NaCl

(b) 42.6g Na3PO4 x 1mol/ 163.9392 Na3PO4 X 1mol/ Ni3(PO4)2/ 2 mol Na3PO4 X 366g Na3PO4/ 1mol Na3(PO4)2

= 43.5 g Ni3(PO4)2

###
Nitric acid HNO3 is manufatured by the Ostwald process in which nitrogen dioxide, NO2, reacts with water.

3NO2 (g) + H2O ---> 2HNO3 + NO

Suppose a vessel contains 4.05 grams NO2 and 6.72 g H20.

(a) What is the limiting reactant?

(b) How many grams of HNO3 could be obtained?

(a) NO2

(b)

NO2-- 4.05g NO2 X 1 mol/ 46.0047g NO2 x 2 mol HNO3/ 3 mol NO2= .05868 = .059

H2O-- 6.72g H2O X 1 mol H2O/ 18.01g H2O X 2 mol HNO3/ 1 mol HNO3 = .07458= .075

Limiting reactant: .059 mol HNO3 X 63.0137 g HNO3/ 1 mol HNO3= 3.7178= 3.72g HNO3

###
Write the oxidation- reduction half reactions for the following reactions and balance the overall reaction.

(a) Mn^2+ + BiO3^- ----> MnO4^- +Bi^3+

(b) Cr2O7^-2 + I^- ---> Cr^3+ + IO3^-

(a) [ Mn^ 2+ ===> Mn^7+ + 5e- ]= 2Mn ---> 2MnO4 + 10 e-

---> 5[ 2e- +BiO3^- ===> Bi^3+] = 70 e- + 5BiO3---> 5Bi

*---> 2Mn + 5BiO3 ---> 5Bi + 2MnO4

(b) [ 6e- + Cr2O7 ---> 2Cr^3+] ===> 6e-+ Cr2)7----> Cr^3+

-----> [I^- (+6) ---> IO3^-] 9 ===> 9I---> 9IO3 + 9e-

**----> 9I +Cr2O7---> Cr+ 9IO3

### A given hydrocarbon is converted completely to water and carbon dioxide, and the mole ratio of H2O to CO2 is 1.33:1.00. The Hydrocarbon could be?

e. C3H8

###
Ammonia can be made by reaction of water with calcium cyanamide:

CaCN2 + 3H2O ---> CaCO3 + 2NH3

When the equation is properly balanced, the sum of the coefficients is:

d.7

###
SnO2 + 2CO --> Sn + 2CO2 DeltaH= +14.7Kj

a. Calculate the amount of heat absorbed when 5.60g of tin are formed.

b. Calculate the volume of CO2 evolved (measured at 745mmHg and 23C) when 5.60 g of tin are formed.

a. 5.60g Sn x 1mol Sn/ 118.71g Sn x 2mol CO2/ 1 mol Sn= .094 mol

PV= nRT ---> V= nRT/P

===> .094 mol (0.0821)(296k)/ (745/760) = 2.2843504/ .98026= 2.33 g/L