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wolverine [178]
3 years ago
7

20 POINTS! SAVE ME PLS!

Chemistry
1 answer:
NISA [10]3 years ago
4 0
ANSWER:
OPTION C) F-1
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The number of significant figures in 0.000820 g is?
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D) releases greenhouse gasses
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All of the above
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8 0
3 years ago
Help Pls!
Salsk061 [2.6K]

Answer:

d= 50.23 g/cm³

Explanation:

Given data:

radius = 137.9 pm

mass is = 5.5 × 10−22 g

density = ?

Solution:

volume of sphere= 4/3π r³

First of all we calculate the volume:

v= 4/3π r3

v= 1.33× 3.14× (137.9)³

v= 1.33 × 3.14 × 2622362.939 pm³

v= 1.095 × 10∧7 pm³

v= 1.095 × 10∧-23 cm³

Formula:

Density:

d=m/v

d= 5.5 × 10−22 g/ 1.095 × 10∧-23 cm³

d= 5.023 × 10∧+1 g/cm³

d= 50.23 g/cm³

8 0
3 years ago
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