Delta H of solution = -Lattice Energy + Hydration
<span>Delta H of solution=- (-730)+(-793) </span>
<span>Delta H of solution= -63kJ/mol </span>
<span>Now we find moles of LiI: </span>
<span>10gLiI/133.85g=.075moles </span>
<span>multiply moles to the delta H of solution to cross cancel moles. .75moles x -64kJ/mol =4.7</span>
(20/23) times 100 is 87%
now the answer asks for percentage ERROR
100-87 =13
so answer is C
Answer:
Explanation:
Gravity pulls the balls down the ramp, and the force of gravity is bigger on larger-mass objects. The extra force on the bigger ball means that it has more energy when it gets to the bottom of the ramp and consequently travels more before stopping.
You should keep the cough drop whole. This maintains the largest surface-to-volume ratio and slows the dissolution of the cough drop.
<h3>Delay of drug action</h3>
The action of the drug can be delayed to either reduce adverse effect if the drugs to the body or to enhance its therapeutic purposes.
Because the throat is dry, the therapeutic purposes of the cough drop would only be achieved if it's dissolution is delayed in the mouth.
To do so, the cough drop should be kept whole to maintain the largest surface-to-volume ratio. This would slow the dissolution of the cough drop.
Learn more about drug here:
brainly.com/question/26254731
Answer:
9.47 mL
Explanation:
The reaction that takes place is:
- 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O
First we <u>calculate how many KOH moles reacted</u>, using <em>the given concentration and volume of KOH solution</em>:
- 0.061 mol/L = 0.061 mmol/mL
- 0.061 mmol/mL * 26.7 mL = 1.6287 mmol KOH
Then we <u>convert KOH moles into H₂SO₄ moles</u>, using the <em>stoichiometric coefficients</em>:
- 1.6287 mmol KOH *
= 0.8144 mmol H₂SO₄
Finally we <u>calculate the required volume of the H₂SO₄ solution</u>, using<em> the number of moles and given concentration</em>:
- 0.8144 mmol ÷ 0.086 mmol/mL = 9.47 mL
