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miss Akunina [59]
2 years ago
5

How many half-lives will pass by the time 1.56% of I-131 is present? B. Approximately how many days does that equal? *

Chemistry
1 answer:
serg [7]2 years ago
6 0

Answer: Hmmmmm that's crazy....

There are a couple of equations one could use for this type of problem, but I find the following to be the easiest to use and to understand.

Fraction remaining (FR) = 0.5n

n = number of half lives that have elapsed

In this problem, we need to find n and are given the FR, which is 1.56% or 0.0156 (as a fraction).

0.0156 = 0.5n

log 0.0156 = n log 0.5

-1.81 = -0.301 n

n = 6.0 half lives have elapsed

Explanation:

Just wanted to help. Hopefully it's correct wouldn't want to waster your time ;)

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1. Using the Slater rule, determine the effective nuclear charge of platinum.
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Answer:

Z* = 3.55

Explanation:

Slater rule says that:

Z*= Z - S

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Z is the nuclear charge

S is the shielding constant

First we write the electronic configuration of platinum:1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6} 4f^{14} 5d^{9} 6s^{1}

The first Slater rule says that we need to group:

(1s^{2}) (2s, 2p)^{8} (3s, 3p)^{8} (3d^{10}) (4s, 4p)^{8} (4d^{10}) (5s, 5p)^{8} (4f^{14}) (5d^{9}) (6s^{1})

The second rule says that the electrons to the right are not shielding, but we are going to solve the exercise for the last level (6s), so we don't have electrons to the right.

For the third rule we have two considerations, if is ns or np and if is nd or nf:

For our case, we have an electro that is in ns, so the rule says that

-electrons within same group shield 0.35, except the 1s which shield 0.30

-electrons within the n-1 group shield 0.85

-electrons within the n-2 or lower groups shield 1.00

Now we can proceed with the calculation:

The first consideration in the third rule does not apply as we only have one electron on this level.

The second consideration will be as follow for the level 5, where we have 17 electrons.

Finally the third consideration will be for levels 1, 2, 3 and 4, where we have 14 for 4f, 10 for 4d, 8 for 4s and 4p, 10 for 3d, 8 for 3s and 3p, 8 for 2s and 2p and finally 2 for 1s, which gives 60 electrons.

So the result for S=(60*1.00 + 17*0.85) = 74.45

And the equation is: Z* = 78 - 74.45

So Z* = 3.55

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2 years ago
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If you look at the combining masses of X and Y in each of the two compounds, 

The first compound contains 0.25g of X combined with 0.75g of Y 
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The second compound contains 0.33 g of X combined with 0.67 g of Y 
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Now, you suppose to prepare each of these two compounds, starting with the same fixed mass of element Y ( I will choose 12g of Y for an easy calculation!) 

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<span>The ratio which combined the masses of X and the fixed mass (12g) of Y
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So, the ratio of MOLES of X which combined with the fixed amount of Y in the two compounds is also = 2 : 3 </span>

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8.04 mL.

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How many moles of NaOH?

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n = c\cdot V = 0.0158 \times0.321 = 0.00507\;\text{mol}.

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