39. calculate the sum of 6.078 g and 0.3329 g.\n40. subtract 7.11 cm from 8.2 cm. (hint: see sample problem…

39. calculate the sum of 6.078 g and 0.3329 g.\n40. subtract 7.11 cm from 8.2 cm. (hint: see sample problem e.)\n41. what is the product of 0.8102 m and 3.44 m?\n42. divide 94.20 g by 3.16722 ml.\n43. write the following numbers in scientific notation.\na. 0.0006730\nb. 50000.0\nc. 0.000003010\n44. the following numbers are in scientific notation. write them in standard notation.\na. 7.050×10³ g\nb. 4.00005×10⁷ mg\nc. 2.3500×10⁴ ml\n45. perform the following operation. express the answer in scientific notation and with the correct number of significant figures. 0.002115 m×0.0000405 m\n46. a sample of a certain material has a mass of 2.03×10⁻³ g. calculate the volume of the sample, given that the density is 9.133×10⁻¹ g/cm³. use the four - step method to solve the problem. (hint: see sample problem f.)

39. calculate the sum of 6.078 g and 0.3329 g.\n40. subtract 7.11 cm from 8.2 cm. (hint: see sample problem e.)\n41. what is the product of 0.8102 m and 3.44 m?\n42. divide 94.20 g by 3.16722 ml.\n43. write the following numbers in scientific notation.\na. 0.0006730\nb. 50000.0\nc. 0.000003010\n44. the following numbers are in scientific notation. write them in standard notation.\na. 7.050×10³ g\nb. 4.00005×10⁷ mg\nc. 2.3500×10⁴ ml\n45. perform the following operation. express the answer in scientific notation and with the correct number of significant figures. 0.002115 m×0.0000405 m\n46. a sample of a certain material has a mass of 2.03×10⁻³ g. calculate the volume of the sample, given that the density is 9.133×10⁻¹ g/cm³. use the four - step method to solve the problem. (hint: see sample problem f.)

Answer

39.

Explanation:

Step1: Add the two masses

$6.078 + 0.3329=6.4109$ g

Answer:

$6.4109$ g

40.

Explanation:

Step1: Subtract the lengths

$8.2 - 7.11 = 1.09$ cm

Answer:

$1.09$ cm

41.

Explanation:

Step1: Multiply the two lengths

$0.8102\times3.44 = 2.787088$ m²

Answer:

$2.787088$ m²

42.

Explanation:

Step1: Divide the mass by the volume

$\frac{94.20}{3.16722}\approx29.74$ g/mL

Answer:

$29.74$ g/mL

43.

Explanation:

a. Move decimal 4 places right

$0.0006730 = 6.730\times10^{-4}$

b. Move decimal 4 places left

$50000.0=5.0000\times 10^{4}$

c. Move decimal 6 places right

$0.000003010 = 3.010\times10^{-6}$

Answer:

a. $6.730\times10^{-4}$ b. $5.0000\times 10^{4}$ c. $3.010\times10^{-6}$

44.

Explanation:

a. Move decimal 3 places right

$7.050\times 10^{3}=7050$ g

b. Move decimal 7 places right

$4.00005\times 10^{7}=40000500$ mg

c. Move decimal 4 places right

$2.3500\times 10^{4}=23500$ mL

Answer:

a. $7050$ g b. $40000500$ mg c. $23500$ mL

45.

Explanation:

Step1: Multiply the numbers

$0.002115\times0.0000405 = 8.56575\times10^{-8}$

Step2: Round to 3 significant figures

$8.57\times10^{-8}$ m²

Answer:

$8.57\times10^{-8}$ m²

46.

Explanation:

Step1: Recall density formula $\rho=\frac{m}{V}$, so $V = \frac{m}{\rho}$

$V=\frac{2.03\times 10^{-3}}{9.133\times 10^{-1}}$

Step2: Divide the numbers

$V=\frac{2.03}{9.133}\times10^{-3 + 1}$ $V\approx0.2223\times10^{-2}$

Step3: Convert to proper scientific - notation

$V = 2.223\times10^{-3}$ cm³

Answer:

$2.223\times10^{-3}$ cm³