Answer:
Complete ionic: .
Net ionic: .
Explanation:
Start by identifying species that exist as ions. In general, such species include:
- Soluble salts.
- Strong acids and strong bases.
All four species in this particular question are salts. However, only three of them are generally soluble in water: , , and . These three salts will exist as ions:
- Each formula unit will exist as one ion and one ion.
- Each formula unit will exist as one ion and two ions (note the subscript in the formula .)
- Each formula unit will exist as one and two ions.
On the other hand, is generally insoluble in water. This salt will not form ions.
Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite , , and (three soluble salts) as the corresponding ions.
Pay attention to the coefficient of each species. For example, indeed each formula unit will exist as only one ion and one ion. However, because the coefficient of in the original equation is two, alone should correspond to two ions and two ions.
Do not rewrite the salt because it is insoluble.
.
Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of and two units of . Doing so will give:
.
Simplify the coefficients:
.
Answer:
1.25 gram of cesium-137 will remain.
Explanation:
Given data:
Half life of cesium-137 = 30 year
Mass of cesium-137 = 5.0 g
Mass remain after 60 years = ?
Solution:
Number of half lives passed = Time elapsed / half life
Number of half lives passed = 60 year / 30 year
Number of half lives passed = 2
At time zero = 5.0 g
At first half life = 5.0 g/2 = 2.5 g
At 2nd half life = 2.5 g/ 2 = 1.25 g
Thus. 1.25 gram of cesium-137 will remain.
Answer:
Option d: C₈H₉NO₂ = acetaminophen, analgesic
Explanation:
% composition of compound is:
63.57 g of C
6 g of H
9.267 g of N
21.17 g of O
First of all we divide each by the molar mass of the element
63.57 g / 12 gmol = 5.29 mol of C
6 g of H / 1 g/mol = 6 mol H
9.267 g of N / 14 g/mol = 0.662 mol of N
21.17 g of O / 16 g/mol = 1.32 mol of O
We divide each by the lowest value, in this case 0.662
5.29 / 0.662 = 8
6 / 0.662 = 9
0.662 / 0.662 = 1
1.32 / 0.662 = 2
Molecular formula of the compound is C₈H₉NO₂
<em><u>Answer and Explanation:</u></em>
<em><u>For % of N2 gas:
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