مواضيع المحاضرة: Quiz I-key-1
قراءة
عرض

KEY

1. Write the number of significant figures in the following measurements:

422 cm 3

6.02 g 3
0.825 m 3
0.0043 g 2
1.310 x 1022 atoms 4
7000 mL 1
0.0000003 cm 3
6001 sec 4

2. Name these:

NaF
Sodium floride
CuClO2
Copper chlorite
Fe2(SO3)3
Iron(III) sulfite
N2O4
Dinitrogen tetroxide
Ca(HCO3)2
Calcium bicarbonate



3. Calculate the percent abundance of two isotopes of a fictional element, georgium, if it has an average atomic mass of 291.23 amu. (Go-290 : 289.86 amu; Go-292 : 292.07 amu)

Percentage of Go-290 =x Percentage of Go-292=y x+y=1

Average atomic mass =291.23 amu = 289.86x +292.07y y=1-x

291.23 =289.86x +292.07(1-x)

(292.07-289.86)x = 292.07-291.23

2.21x=0.84 x=0.38 y=1-x=1-0.38 =0.62

38% Go-290 32% Go-292

4. When 0.115 g of a hydrocarbon is burned, 0.379 g of CO2 and 0.1035 g of H2O are produced. What is the empirical formula of the hydrocarbon?(molecular mass of CO2 = 44 g/mol; H2O=18 g/mol)
Hydrocarbon CxHy

nC = 0.379g CO2 x 1 mol CO2 /44 g x 1mol C /1 mol CO2 =0.09

nH = 0.135g H2O x 1 mol H2O / 18g x 2mol H / 1mol H2O = 0.12


C0.09 H0.12 both parts divided by 0.09 CH1.33 multiplied by 3 C3H4

5. Express the following numbers in scientific notation:

0.00000000066 6.6 x10-10
356 3.56 x102
47,764 4.78 x 10
0.096 9.6 x 10-2

6. Indicate the number of protons, neutrons and electrons in an element with atomic number of 38 and mass number of 84.

Atomic number = numbers of protons =38

and if the element has no charge = numbers of electrons= numbers of protons =38

Mass number = numbers of protons+ numbers of neutrons

84 = 38 + numbers of neutrons

numbers of neutrons = 84-38 = 52

7. The density of ammonia gas under certain conditions is 0.625 g/L. Calculate its density in g/cm3.


0.625 g/L x 1L/1dm3 x 1 dm3/1000 cm3 = 6.25 x 10-4 g/cm3

8. Carry out the following arithmetic operations to the correct number of significant figures:

11,254.1 g + 0.1983 g = 11,254.3 g

66.59 L – 3.113 L = 63.48 L

8.16 m x 5.1355 m =41.9 m

0.0154 kg / 88.3 ml =0.000174 mL

(2.64 x 103 cm) + (3.27 x 102 cm) = 272 + 334 = 606 cm

7.310 km / 5.70 km = 1.28 km



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