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zhuklara [117]
3 years ago
9

Assume a reaction takes place in a basic solution to form the given products: mno4-(aq) + cl-(aq) --------> mno2(s) + cl2(g)

(unbalanced)
i. Balance the given half reactions for atoms and charge
Chemistry
1 answer:
VARVARA [1.3K]3 years ago
3 0
<span>To balance, we write it in its half reactions and balance the half reactions by adding H2O, OH-, H+ and electrons. We do as follows:

MnO4-(aq) + Cl-(aq) --------> MnO2(s) + Cl2(g) 


</span>2 (2H2O + MnO2 --> MnO4- + 4H+ 3e- )
<span>3 (2e- + Cl2 --> 2Cl- )
</span>
4H2O + 2MnO2 -----> 2MnO4- + 8H+ + 6e- 
<span>6e- + 3Cl2 -------> 6Cl- 
</span>----------------------------------------------------------------------
<span>4H2O + 2MnO2 + 3Cl2 --------> 2MnO4- + 6Cl- + 8H+ 

Since it is in a basic solution,

</span>8OH- + 2MnO2 + 3Cl2 -----> 2MnO4- + 6Cl- + 4H2O 



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6 0
3 years ago
Read 2 more answers
If the half-life of C-14 is 5730 years, how much C-14 would be left from a 5000 year-old sample that started as 150 grams?
serious [3.7K]
<h3>Answer:</h3>

81.9 grams

<h3>Explanation:</h3>

From the question we are given;

  • Half-life of C-14 is 5730 years
  • Original mass of C-14 (N₀) = 150 grams
  • Time taken, t = 5000 years

We are required to determine the mass left after 5000 years

  • Using the formula;
  • N = No(1/2)^t/T​, where N is the remaining mass, N₀ is the original mass, t is the time taken and T is the half-life.

   t/T = 5000 yrs ÷ 5730 yrs

    = 0.873

N = 150 g ÷ 0.5^0.873

   = 150 g × 0.546

   = 81.9 g

Therefore, the mass of C-14 left after 5000 yrs is 81.9 g

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