Answer:
Terms in this set (11)
Na. Sodium / Natrium.
K. Potassium / Kalium.
Fe. Iron / Ferrum.
Cu. Copper / Cuprum.
Sb. Antimony / Stibium.
Au. Gold / Aurum.
Pb. Lead / Plumbum.
Hg. Mercury / Hydragyrum.
Answer : The new pressure acting on a 2.5 L balloon is, 8.6 atm.
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.
or,
where,
= initial pressure = 3.7 atm
= final pressure = ?
= initial volume = 5.8 L
= final volume = 2.5 L
Now put all the given values in the above equation, we get:
Thus, the new pressure acting on a 2.5 L balloon is, 8.6 atm.
Answer:
5.916x10⁻³ mol OH⁻
Explanation:
The reaction that takes place is:
- H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
First we <u>calculate the added moles of each reagent</u>, using the <em>given volumes and concentrations</em>:
- H₂SO₄ ⇒ 0.144 M * 27.55 mL = 3.967 mmol H₂SO₄
- KOH ⇒ 0.316 M * 43.84 mL = 13.85 mmol KOH
Now we<u> calculate how many KOH moles reacted with 3.967 mmol H₂SO₄</u>:
- 3.967 mmol H₂SO₄ * = 7.934 mmol KOH
Finally we calculate how many OH⁻ moles remained after the reaction
- 13.85 mmol - 7.934 mmol = 5.916 mmol OH⁻
- 5.916 mmol / 1000 = 5.916x10⁻³ mol OH⁻
- Iron is in limited amount, limiting reagent
- The theoretical yield of iron(III) oxide formed is 27.1 grams.
Given:
The reaction between iron and dioxygen at high temperatures to form iron(III) oxide.
To find:
- The limiting reagent
- The theoretical yield of iron(III) oxide formed when Suppose 19.1 g
of iron is reacted with 17.9 g of oxygen.
Solution:
- The mass of iron = 19.1 g
- The moles of iron
- The mass of dioxygen = 19.1 g
- The moles of dioxygen
According to reaction, 4 moles of iron reacts with 3 moles of dioxygen then 0.342 moles of iron will react with:
0.2465 moles of dioxygen will react with 0.342 moles of iron which means that:
- Iron is in limited amount, limiting reagent
- dioxygen is in an excessive amount, excessive reagent.
- The amount of iron(III) oxide formed will depend upon moles of iron.
According to reaction, 4 moles of iron gives 2 moles of iron (III) oxide, then 0.342 moles of iron will give:
The theoretical yield of 0.171 moles of iron(III) oxide:
The theoretical yield of iron(III) oxide formed is 27.1 grams.
Learn more about limiting and excessive reagent here:
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