Do 3&4 Express each number in scientific notation. a. 1, 650, 763, 73 wavelength

Question

Do 3&4 Express each number in scientific notation. a. 1, 650, 763, 73 wavelengths ___________ b. 0.0000000181 cm ___________ c. 0.0056 times 10^-3 ___________ Solve the following problems and express the answer in exponential the notation. Include proper number of significant figures in your answer. a. (0.08)(5.44 times 10^-3)(0.00467)/(87.33)(22.45 times 10^15)(0.0356) ___________ b. 1.080 + 3.000 + 0.02 + 9.00463 ___________ Two sheets of metal, one of which was known to be iron and the other nickel, were accidentally separated and the identity of each was lost. The sheets were labeled A and B, and both were weighed. Sheet A weighed 9.48g and sheet B weighed 10.89g. When A was placed in 50.00 mL of water, the total volume was 51.2 mL. When B was placed in 50.00 mL of water, the volume was 51.1 mL. The density of iron is 7.9 g/mL and that of nickel is 9.9 g/mL. Calculate the density of each metal and identify each sheet of metal. Metal A = ___________ Metal B = ___________ Aluminum metal has a density of 2.7 g/cm^3. a. Calculate the mass, in g, of an aluminum sample with a volume of 60.0 cm^3. ___________ b. b. Calculate the volume, in LITERS, of an aluminum sample that has a mass of 98.5 g. ___________

Explanation / Answer

3. We know, Density = Mass/Volume

in this part we are given the following information

Volume of metal=

(V2-V1)

CALCULATIONS;

If we subtract the total volume and volume of water in which metal was placed we will get volume of metal, which we would be needing in our formula of density .

Metal A - Volume of metal = 51.2-50.0 = 1.2mL;

Density = 9.48g/1.2mL = 7.9g/mL

and if we match with the density of metals given in the question, we find that this is the density of iron. So we conclude that metal A is IRON.

Metal B - Volume of metal = 51.1-50.0= 1.1mL

Density = 10.89g/1.1mL = 9.9g/mL

and if we match with the density of metals given in the question, we find that this is the density of nickel. So we conclude that metal B is NICKEL.

4. Density of aluminium 2.7g/cm3 and Density = Mass/Volume

a. We are given volume = 60.0cm3 and we need to find mass in g,

using above formula; 2.7g/cm3 = Mass/60.0cm3

Mass = 2.7g/cm3 x 60.0cm3 = 162.0g

b. We are given mass = 98.5g and we need to find volume in LITRES

For this we need to know the unit conversion that 1 Litre = 1000cm3. Now solving further

2.7g/cm3 = 98.5g/Volume

Volume =98.5g/2.7g/cm3 = 36.5cm3  

Now calculating cm3 into litres we get that 36.5cm3 is 36.5x10-3Litres.

Metal Mass(g) Volume of water/V1 Total volume/V2

Volume of metal=

(V2-V1)

Density A 9.48 50.0 51.2 1.2 7.9 B 10.89 50.0 51.1 1.1 9.9

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