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To answer this question, you need to know the concept of half-life, which is how a radioactive material decreases in mass over time.
The half life of U-235 is 703.8 million years. The first part of this problem is to find the scale factor. To do this, divide the time that has past by the half life, like this:
Now, take this scale factor and multiply it by the current mass, like this:
This number is what you add to the current mass to get the original mass. That is because the scale factor showed us that it was just over one half life. Since after one half life, the mass is cut in half, and this is over one half life, when we add to the original it will be a little over double. This equation illustrates the final addition:
I hope this helped you. Fell free to ask any further questions.
The half life of U-235 is 703.8 million years. The first part of this problem is to find the scale factor. To do this, divide the time that has past by the half life, like this:
Now, take this scale factor and multiply it by the current mass, like this:
This number is what you add to the current mass to get the original mass. That is because the scale factor showed us that it was just over one half life. Since after one half life, the mass is cut in half, and this is over one half life, when we add to the original it will be a little over double. This equation illustrates the final addition:
I hope this helped you. Fell free to ask any further questions.
The formal charges of all nonhydrogen atoms are -1.
Solution:-
<u>O 7-4 = 3 O Double bond on one H 5-4 = 1</u>
O-Cl-O 6-7 = -1x4 = -4 N 5-4=1 H-N-H 1-1=0
O 3-4= -1 O O 6-7 = -1(2)=-2 H 1-0=+1
<u>6-6 = 0 1-2 = -1</u>
It will percentage its last valence electron thru a single bond to the terminal oxygen atom. This is in agreement with carbon and hydrogen atoms that each need to form 4 and 1 covalent bonds respectively. because the terminal oxygen atom best has a single covalent bond, it'll have a proper rate of -1.
According to the lewis structure of SO2, The critical atom is sulfur and it is bonded with 2 oxygen atoms thru a double bond. each oxygen atom acquires 2 lone pairs of electrons and the primary sulfur atom has 1 lone pair of electrons.
Learn more about Nonhydrogen atoms here:-brainly.com/question/2822744
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212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Option A.
Explanation:
Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.
So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.
Then,
So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.