Answer:
(a) 1.95 × 10⁴ g
(b) 1.95 × 10⁷ mg
(c) 1.95 × 10¹⁰ μg
Explanation:
A dog has a mass of 19.5 kg.
<em>(a) What is the dog's mass in grams?</em>
1 kilogram is equal to 10³ grams. The mass of the dog in grams is:
19.5 kg × (10³ g/1 kg) = 1.95 × 10⁴ g
<em>(b) What is the dog's mass in milligrams?</em>
1 gram is equal to 10³ milligrams. The mass of the dog in milligrams is:
1.95 × 10⁴ g × (10³ mg/1 g) = 1.95 × 10⁷ mg
<em>(c) What is the dog's mass in micrograms?</em>
1 gram is equal to 10⁶ micrograms. The mass of the dog in micrograms is:
1.95 × 10⁴ g × (10⁶ μg/1 g) = 1.95 × 10¹⁰ μg
Answer:
15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.
Explanation:
A → B + C
The rate law of the reaction will be :
Initial rate of the reaction when concentration of the substrate was 0.4 M:
..[1]
Initial rate of the reaction when concentration of the substrate was 0.8 M:
...[2]
[1] ÷ [2] :
x = 0
The order of the reaction is zero.
For the value of rate constant ,k:
..[1]
x = 0
k= 0.183 M/s
The half life of the zero order kinetics is given by :
Where:
= Initial concentration of A
k = Rate constant of the reaction
So, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M:
15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.
Answer:
4.3x10⁻⁵ mol/L.
Explanation:
Hello there!
In this case, since the dissolution of silver phosphate is not completely achieved due to its low solubility in water, we can represent this as an equilibrium problem by which it separates into silver and phosphate ions:
Which are slightly soluble in water; it means that the equilibrium expression is:
Thus, since we know the solubility product and the concentration of these ions in a saturated solution is represented by the reaction extent , we can write:
Thus, solving for we obtain:
Thus, the molar concentration of silver ions is 4.3x10⁻⁵ mol/L.
Best regards!