Answer:
4KO₂ + 2CO₂ -> 2K₂CO₃ + 3O₂
<u> Step 1: Find the moles of O₂.</u>
n(O₂) = mass/ Mr.
n(O₂) = 100 / 32 = 3.125 mol
<u>Step 2: Find the ratio between KO₂ and O₂.</u>
<u>KO₂ </u> : <u> O₂</u>
4 : 3
4/3 : 1
(4*3125)/3 : 3.125
=4.167 mol of KO₂
Thus now we know, to produce 100 g of O₂, we need 4.167mol of KO₂
<u>Step 3: Find the mass of KO₂:</u>
<u />
mass = mol * Mr. (KO₂)
Mass = 4.167* 71.1
Mass = 296.25 g
Hello!
When HI is added, the buffer reacts in the following way:
1) Neutralizing of the Acid:
HI + NH₄OH → NH₄I + H₂O
2) Dissociation of the salt of a weak acid:
NH₄I → NH₄⁺(aq) + I⁻ (aq)
3) Dissociation of a weak acid to form H₃O⁺ (very little):
NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺
This series of reactions show how the adding of a strong Acid can be neutralized by the buffer, releasing instead very little amounts of Hydronium ions.
Have a nice day!
Answer:
d ≈ 0.098 g/mL
Explanation:
The density of a substance can be found by dividing the mass by the volume.
d=m/v
The mass of the substance is 0.221 grams and the volume is 2.25 milliliters.
m= 0.221 g
v= 2.25 mL
Substitute the values into the formula.
d= 0.221 g / 2.25 mL
Divide
d= 0.098222222 g/mL
Let’s round to the nearest thousandth. The 2 in the ten thousandths tells us to keep the 8 in the thousandth place.
d ≈ 0.098 g/mL
The density of the substance is about 0.098 grams per milliliter.
The chemical reaction would be:
<span>CuSO4(aq) + NaOH(aq) = Cu(OH)2(s) + Na2SO4(aq)
One observation would be the formation of a precipitate since one of the products is not readily soluble to aqueous solution. A formation of a blue precipitate will be observed.</span>
