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Serhud [2]
3 years ago
8

Cobalt-60 and iodine-131 are radioactive isotopes that are used in

Chemistry
1 answer:
svlad2 [7]3 years ago
8 0
Cobalt 60-Used to treat certain types of cancer.
Iodine-131-Used to monitor and treat goiter and other thyroid problems.It is also used to treat liver and brain tumor
Mark me as the Brainliest❤️❤️✌✌
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N76 [4]

Answer:

b ella how does any sond 22md how 22x 22x 2md rose blackpink hottest robot

6 0
3 years ago
if an endothermic reaction begins at 26°C and decreases by 2°C per minute how long will it take to reach 0°C?
KATRIN_1 [288]

Answer:

13 minutes

Explanation:

For the endothermic reaction to reach 0°C, it will take 13 minutes.

 Let us follow the process step by step;

 Rate of decrease is 2°C per minute.

  Start is 26°C

    Time                   temperature

     0 min                    26°C

      1 min                     24°C

      2 min                     22°C

     3 min                      20°C

    4 min                       18°C

   5 min                        16°C

     6 min                      14°C

     7 min                      12°C

     8 min                     10°C

    9 min                       8°C

     10 min                      6°C

     11 min                       4°C

     12 min                      2°C

      13 min                      0°C

4 0
3 years ago
Determine the [OH-], pH, and pOH of a solution with a [H+] of 0.090 M at 25 °C. [OH-] = pH = pOH = 0 POH =
ValentinkaMS [17]

Answer : The concentration of OH^- ion, pH and pOH of solution is, 1.12\times 10^{-13}M, 1.05 and 12.95 respectively.

Explanation : Given,

Concentration of H^+ ion = 0.090 M

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

The expression used for pH is:

pH=-\log [H^+]

First we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.090)

pH=1.05

The pH of the solution is, 1.05

Now we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-1.05=12.95

The pOH of the solution is, 12.95

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

12.95=-\log [OH^-]

[OH^-]=1.12\times 10^{-13}M

The OH^- concentration is, 1.12\times 10^{-13}M

5 0
3 years ago
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irina [24]

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65

Explanation:

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