Answer:
Explanatio(3) Atomic masses are a weighted average of the naturally occurring isotopes.
(3) Atomic masses are a weighted average of the naturally occurring isotopes. is the best statement for why most atomic masses on the Periodic Table are decimal numbers.
The density of the gold is 19.3 grams/cc so each cc weighs 19.3grams. Now we can obtain the volume of gold from the given dimensions ie 4.72x8.21x3.98= 154.23 cc. So for the answer, just multiply the volume or 154.23 x 19.3= 2976.6 grams is the answer.
The pressure of 2.29 atm can be converted to 
<h3>What is conversation?</h3>
Conversation is a way of writing a value in another unit, it helps to reduce large values by using a unit.
It should be noted that 1 atm. =760.0 mm Hg
We were given 2.29 atm, the we can convert it to mm Hg. as;
![[ 2.29 atm* 760.0 mmHg / atm.]](https://tex.z-dn.net/?f=%5B%202.29%20atm%2A%20760.0%20mmHg%20%2F%20atm.%5D)
= 1740.4 mm Hg.
Therefore, The pressure of 2.29 atm can be converted to 
Learn more about conversation at:
brainly.com/question/5962406
Answer:

Explanation:
Na₂S₂O₃ solution does not react with KI (it reacts with I₂), so it is simply diluting the KI, and we can use the dilution formula.

Data:
c₁ = 0.15 mol·L⁻¹; V₁ = 5 drops
V(Na₂S₂O₃) = 40 drops
Calculations:
(a) Calculate the total volume
V₂ = 5 + 40 = 45 drops
(b) Calculate the concentration
0.15 × 5 = c₂ × 45
0.75 = 45c₂
