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Roman55 [17]
3 years ago
13

4.564000 x 10^-4 express how many significant figures

Chemistry
1 answer:
Vaselesa [24]3 years ago
5 0

Answer: 7

Explanation:

Before a number but after a decimal. The zeros at the end would usually mean that it doesn't count but since the numbers are before the zeros and after a decimal it's 7 sig figs

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At the center of every chlorophyll molecule, in every plant, there is a magnesium ion. Magnesium is one of the two dozen or so elements that are <em>required </em>for life. 
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At 40 degrees Celsius , 100 grams of NaClO3 is added to 100 grams of water. The solution would be ..
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Converts chemical energy into mechanical energy
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4 0
2 years ago
Potassium is isotopic and has RAM of 39.5 work out the percentage abundance of each isotope in a given sample of potassium which
Troyanec [42]

Yo sup??

Let the percentage of K-39 be x

then the percentage of K-40 is 100-(x+0.01)

We know that the net weight should be 39.5. Therefore we can say

(39*x+40*(100-(x+0.01))+38*0.01)/100=39.5  

(since we are taking it in percent)

39*x+40*(100-(x+0.01))+38*0.01=3950

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Hope this helps.

3 0
3 years ago
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


4 0
3 years ago
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