Chem Final I

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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

b. 6
Chap. 1

d. 1,2,3
Chap. 1

d. 5
Chap. 1

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

c. 7.4 g/cm^3

Chap. 1

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

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

e. NdN

Chap. 1

b.2

b. 2

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

d. 50.0

e. C3H8

b. 25.0g CrO2

d.7

e. HNO3

a. HClO2

b. H3PO4

e. +3

Which of the following reactions is an oxidation-reduction reaction?

b. NH4NO3 ---> N2O + 2H2O

e. reduction
END of 3&4

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

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