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Zinaida [17]
3 years ago
9

A radioisotope has a half-life of 12 days. If a sample is measured to contain 5 grams of radioisotope, how much time has passed

if the sample originally measured 160 grams of radioisotope?​
Chemistry
1 answer:
Sophie [7]3 years ago
7 0
60 days have passed. In every 12 days the radioisotope loses half of its life. So in the first 12 days, from 160 grams it decreases to half so it becomes 80 grams. The second 12 days, from 80 grams, it decreases to half and becomes 40 g. The third 12 days, from 40 it becomes half to 20 g. The fourth 12 days, from 20 g it becomes half to 10 g. And in the fifth 12 days, from 10 g it becomes half to 5 grams. So 5 times it became half.
5 x 12 ( days)= 60 days
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The atom theory has been around since the 19th centry 
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If an atom of nitrogen bonded with an atom of aluminum , what kind of bond would form
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A/1 * B/C =D
Novay_Z [31]
<h3>Answer:</h3>

                  0.64 Moles of Propane

<h3>Explanation:</h3>

Data:

         Moles of Carbon  =  1.5 mol

         Conversion factor  =  7 mol C produces = 3 mol of Propane

Solution:

             As we know,

                7 moles of Carbon produces =  3 moles of Propane

Then,

            1.5 moles of Carbon will produce  =  X moles of Propane

Solving for X,

                     X =  (1.5 moles × 3 moles) ÷ 7 moles

                     X  =  0.6428571 moles of Propane

Or rounded to two significant figures,

                     X =  0.64 Moles of Propane

5 0
3 years ago
Name the following compound:<br> CH3-CH2-CH2-CH2-CH3<br> CH3 CH3
stepan [7]

Answer:

<u><em>Pentane </em></u>

Explanation:

since we have in here CH3-CH2-CH2-CH2-CH3 5 Carbon atoms and 12 Hydrogen making it C_{5} H_{12}

6 0
3 years ago
If 14.5 g of MnO4- (permanganate) react with manganese (II) hydroxide how many grams of manganese (IV) oxide will be produced? T
Veronika [31]

Answer:

m_{MnO_2}=21.2gMnO_2

Explanation:

Hello,

In this case, given the balanced reaction:

2MnO_4^-+2Mn(OH)_2\rightarrow 4MnO_2+2OH^-+H_2O

We can see a 2:4 mole ration between permanganate ion (118.9 g/mol) and manganese (IV) oxide (86.9 g/mol), that is why the resulting mas of this last one turns out:

m_{MnO_2}=14.5gMnO_4^-*\frac{1molMnO_4^-}{118.9gMnO_4^-}*\frac{4MnO_2}{2molMnO_4^-}  *\frac{86.9gMnO_2}{1molMnO_2} \\\\m_{MnO_2}=21.2gMnO_2

Best regards.

5 0
4 years ago
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