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nalin [4]
3 years ago
14

What is the empirical formula for the compound P4O6??

Chemistry
1 answer:
777dan777 [17]3 years ago
5 0
The empirical formula is P₂O₃
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Draw the structure(s) of all of the possible monochloro derivatives of 2,4-dimethylpentane, c7h15cl.
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The monochloroderivatives will be obtained by substituting chemically non equivalent hydrogen with chlorine atom, one by one

So the possible monochloro derivatives of 2,4-dimethylpentane (figure 1) are shown in figure (2)




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Can someone help me here please i need this :(​
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Start by adding the numbers then divide
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How do you find the scientific notation of a Avogadro's number??
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Answer: ummmmmm hmmmmm

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Is the maximum population that a given area can support
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I took the test

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Read 2 more answers
1. Which of the following nuclear equations is correctly balanced? (choices attached)
lawyer [7]

1. <em>Balancing nuclear equations </em>

Answer:

_{18}^{37}\text{Ar} + _{-1}^{0}\text{e} \rightarrow _{17}^{37}\text{Cl}

Explanation:

The main point to remember in balancing nuclear equations is that the <em>sums of the superscripts</em> (the mass numbers) and the <em>subscripts</em> (the nuclear charges) <em>must balance</em>.

Mass numbers: 37 + 0 = 37; balanced.

Charges: 18 + 1 = 17; balanced

B is <em>wrong</em>. Mass numbers not balanced. 6 +2(1) ≠ 4 + 3.

C is <em>wron</em>g. Mass numbers not balanced. 254 + 4 ≠ 258 + 2(1).

D is <em>wron</em>g. Mass numbers not balanced. 14 + 4 ≠ 17 + 2.

===============

2. <em>Amount remaining </em>

Answer:

D. 5.25 g

Explanation:

The half-life of Th-234 (24 da) is the time it takes for half the Th to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.

We can construct a table as follows:  

  No. of                  Fraction       Amount  

<u>half-lives</u>   <u>t/(da</u>)  <u>remaining</u>  <u>remaining/g</u>  

      1             24            ½                21.0  

      2            48            ¼                10.5  

      3             72           ⅛                 5.25  

      4             96          ¹/₁₆                 2.62  

We see that 72 da is three half-lives, and the amount of Th-234 remaining is 5.25 g.  

===============

3.<em> Calculating the half-life </em>

Answer:

a. 2.6 min

Explanation:

The fraction of the original mass remaining is 1.0 g/4.0 g ≈ ¼.

We saw from the previous table that it takes two half-lives to decay to ¼ of the original amount.

2 half-lives = 5.2 min       Divide both sides by 2

  1 half-life = 5.2 min/2 = 2.6 min

7 0
3 years ago
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