3.47 x
atoms of gold have mass of 113.44 grams.
Explanation:
Data given:
number of atoms of gold = 3.47 x
mass of the gold in given number of atoms = ?
atomic mass of gold =196.96 grams/mole
Avagadro's number = 6.022 X 
from the relation,
1 mole of element contains 6.022 x
atoms.
so no of moles of gold given = 
0.57 moles of gold.
from the relation:
number of moles = 
rearranging the equation,
mass = number of moles x atomic mass
mass = 0.57 x 196.96
mass = 113.44 grams
thus, 3.47 x
atoms of gold have mass of 113.44 grams
Answer:
3.4752 moles of water
Explanation:
There are 13.84 mole in one cup of water so,
13.84 divided by 4= 3.4725 :)
The mass of plutonium that will remain after 1000 years if the initial amount is 5 g when the half life of plutonium-239 (239pu, pu-239) is 24,100 years is 2.5 g
The equation is Mr=Mi(1/2)^n
where n is the number of half-lives
Mr is the mass remaining after n half lives
Mi is the initial mass of the sample
To find n, the number of half-lives, divide the total time 1000 by the time of the half-life(24,100)
n=1000/24100=0.0414
So Mr=5x(1/2)^1=2.5 g
The mass remaining is 2.5 g
- The half life is the time in which the concentration of a substance decreases to half of the initial value.
Learn more about half life at:
brainly.com/question/24710827
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<em><u>Question</u></em>
<em><u>What </u></em><em><u>does </u></em><em><u>it </u></em><em><u>mean </u></em><em><u>to </u></em><em><u>optimize</u></em><em><u> </u></em><em><u>a </u></em><em><u>solution?</u></em>
<em><u>To find out best possible solution for a given problem within the given constraint is generally termed as optimization</u></em>
<em><u>How </u></em><em><u>are </u></em><em><u>solution</u></em><em><u> </u></em><em><u>optimize</u></em><em><u> </u></em><em><u>?</u></em>
<em><u>To solve an optimization problem, begin by drawing a picture and introducing variables. Find an equation relating the variables. Find a function of one variable to describe the quantity that is to be minimized or maximized. Look for critical points to locate local extrema.</u></em>