7edf-4900-8133-09a99120fa82 CHEM 1F92 Assignment 1. Matter and Measurement Assig

Question

7edf-4900-8133-09a99120fa82 CHEM 1F92 Assignment 1. Matter and Measurement Assignment 1, Version #62 Report all numerical answers to the correct number of significant figures! 1. (a) Give the names of the following elements: @)Li (i) Cr (ii) zinc (b) Give the symbols of the following elements: (i) fluorine 2. Report the answers of these caleulations to the correct mumber of significant figures: (a) 1.5413 x 6.205 x 4.53 (b)36.033+4.0316+159.808-? (e)7 3128 x 10 x5.645 x 10)/4848- (Report answer in exponential notation.) 3. A sheet of metal is 28.8 mm wide and 93.9 mm long. If it weighs 11.17 g and the density of the metal is 8.96 g/em, what is the thickness of the sheet (in mm)? 4. IrPV-eR(T-273.15)M, solve for M when P-451, V-0.120, g-0568, R-62.37, and T-34 An ore contains 40,8 % ofthe mineral magnesite, MgCO,, which is a source ofthe element Mg. How much ore must be processed in order to obtain 15.0 kg of Mg? 6. A piece of iron wire has a diameter of 0.426 mm. If irom has a density of 7.86 glee, how long (in meters) should you cut a picce of wire to obtain 0.0187 moles of iron? 7. Suppose you make a solution that contains 20.33 g of sodium chloride and 371 g of water. (a) what is the mass % of sodium chloride? (b) If 10.22 mL of your solution weighs 10.583 g.what is the density (in gloc)? 8. A crystal of topaz has a density of 3.53 g/cc. (a) What is the mass (in grams) of a topaz crystal with a volume of 1.893 cm (b) What is the density of topaz in kg'm? (c) What is the volume (in mL) of a topaz crystal that weighs 3050 mg? 9. The density of a solution of potassium oxalate is 1.1057 gloc, and it is 14.00 percent by mass. What volume of the solution (in mallaliters) do you need to supply 22.0 g of potassium oxalate?

Explanation / Answer

1.

a)

l) Name of the element Li is : Lithium

ll) Name of the element Cr is : Chromium

b)

l) Symbol of the element fluorine is : F

ll) Symbol of the element Zinc is : Zn

2.

a) 43.324

b)199.873

c) 8.515×10^-8

4.

M =[GR(T+273.15)]÷PV

M=201.06

5.

( 15×100)/ 40.8

= 36.76kg ore required

7.

a) mass= w/v = 20.33/371

= 0.0548

b)Density = w/v = 10.583/10.22

=0.966gr/cc

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.