Answer:
1Au(s) + 3HNO₃(aq) + 4HCl(aq) → 1HAuCl₄(aq) + 3NO₂(g) + 3H₂O(l)
The function of HCl is oxidize the gold.
Explanation:
It is possible to balance the reaction seeing each compound as a variable and take an equation for each atom, thus:
Au(s) + HNO₃(aq) + HCl(aq) → HAuCl₄(aq) + NO₂(g) + H₂O(l)
a + b + c = d + e + f
Au: a = d <em>(1)</em>
H: b + c = d + 2f <em>(2)</em>
N: b = e <em>(3)</em>
O: 3b = 2e + f <em>(4)</em>
Cl = c = 4d <em>(5)</em>
Assuming <em><u>a = 1</u></em>:
<em><u>1 = d</u></em> <em>(1)</em>
<u><em>c = 4</em></u> <em>(5)</em>
b + 3 = 2f <em>(2)</em>
3b = 2e + f <em>(4)</em>
As b = e:
b = f <em>(4)</em>
<em><u>f = 3</u></em>; <em><u>b = 3</u></em>; <em><u>e = 3</u></em>
Thus, balanced reaction is:
1Au(s) + 3HNO₃(aq) + 4HCl(aq) → 1HAuCl₄(aq) + 3NO₂(g) + 3H₂O(l)
<em>The function of HCl is oxidize the gold</em> that before reaction is Au⁰ and afte ris Au⁺³
I hope it helps!
Physical chemistry.
organic chemistry.
inorganic chemistry.
analytical chemistry.
biochemistry.
Answer:
Half life of phosphorous-32 = 14 days
Explanation:
Given data:
Total mass of phosphorous-32 = 2.0 g
After 42 days mass left = 0.25 g
Half life of phosphorous-32 = ?
Solution:
First of all we will calculate the number of half lives passed.
At time zero = 2.0 g
At first half life = 2.0 g/2 = 1.0 g
At 2nd half life = 1.0 g/2 = 0.5 g
At 3rd half life = 0.5 g/2 = 0.25 g
Half life:
Half life = T elapsed / half lives
Half life = 42 days/ 3
Half life = 14 days
