Question

instructions: silently and independently, answer the following equations with the correct significant figures and units.

1. 3.45 g + 1.231 g + 1.2 g=

#of sig figs:
answer:
final answer:

1. 6 g + 0.5 g + 100 g=

#of sig figs:
answer:
final answer:

1. 1000.00 g + 1200.001 g + 100.12 g=

#of sig figs:
answer:
final answer:

1. 4.00 ml x 2.1 ml x 1.111 ml=

#of sig figs:
answer:
final answer:

1. 5 ml x 2.1 ml x 3.333 ml=

#of sig figs:
answer:
final answer:

1. 100.00 ml x 100.1 ml x 1 ml=

#of sig figs:
answer:
final answer:

Explanation:

Step1: Add the masses for the first addition problem

$3.45\ g+1.231\ g + 1.2\ g=5.881\ g$. For addition, we round to the least number of decimal - places among the values being added. 1.2 has the least number of decimal places (1 decimal place). So, $5.881\ g\approx5.9\ g$. The number of significant figures is 2.

Step2: Add the masses for the second addition problem

$6\ g+0.5\ g + 100\ g=106.5\ g$. 6 and 100 have the least precision (no decimal places). So, $106.5\ g\approx107\ g$. The number of significant figures is 3.

Step3: Add the masses for the third addition problem

$1000.00\ g+1200.001\ g+100.12\ g = 2300.121\ g$. 1000.00 has the least number of decimal places among them. Rounding to two decimal places, we get $2300.12\ g$. The number of significant figures is 6.

Step4: Multiply the volumes for the fourth multiplication problem

$4.00\ mL\times2.1\ mL\times1.111\ mL=9.3324\ mL^{3}$. For multiplication, we round to the least number of significant figures among the values being multiplied. 2.1 has 2 significant figures. So, $9.3324\ mL^{3}\approx9.3\ mL^{3}$. The number of significant figures is 2.

Step5: Multiply the volumes for the fifth multiplication problem

$5\ mL\times2.1\ mL\times3.333\ mL = 34.9965\ mL^{3}$. 5 has 1 significant figure. So, $34.9965\ mL^{3}\approx30\ mL^{3}$. The number of significant figures is 1.

Step6: Multiply the volumes for the sixth multiplication problem

$100.00\ mL\times100.1\ mL\times1\ mL=10010\ mL^{3}$. 1 has 1 significant figure. So, $10010\ mL^{3}\approx10000\ mL^{3}$. The number of significant figures is 1.

Answer:

1. #of sig figs: 2, Final answer: $5.9\ g$
2. #of sig figs: 3, Final answer: $107\ g$
3. #of sig figs: 6, Final answer: $2300.12\ g$
4. #of sig figs: 2, Final answer: $9.3\ mL^{3}$
5. #of sig figs: 1, Final answer: $30\ mL^{3}$
6. #of sig figs: 1, Final answer: $10000\ mL^{3}$