Answer:
Electrolytes are substances that can ionize in water. They could be acids, bases or salts as long as they give ions when they dissolve in water.
Explanation:
- <em>Strong electrolytes</em> completely ionize when dissolved in water, leaving no neutral molecules. The strong electrolytes here are:<u> salt water</u>, <u>baking soda (NaHCO3) solution.</u>
- <em>Weak electrolytes</em> do not completely dissociate in solution, and hence have a low ionic yield. Examples of this would be<u> vinegar </u>and <u>bleach </u>(which could be sodium hypochlorite or chlorine, which are weakly dissociated).
- <em>Non-electrolytes </em>will remain as molecules and are not ionized in water at all. In this case, <u>sugar solution is a non-electrolytes</u>, even though sugar dissolves in water, but it remains as a whole molecule and not ions.
Answer:
The correct answer is - yes, 4.57 g of solute per 100 ml of solution
Explanation:
The correct answer is yes we can calculate the solubility of X in the water at 22.0°C. The salt will remain after the evaporate from the dissolved and cooled down at 26°C.
Then, the amount of solute dissolved in the 700 ml solution at 26°C is the weighed precipitate: 0.032 kg = 32 g.
Then solublity will be :
32. g solute / 700 ml solution = y / 100 ml solution
⇒ y = 32. g solute × 100 ml solution / 700 ml solution = 4.57 g.
Thus, the answer is 4.57 g of solute per 100 ml of solution.
Answer:
Explanation:
We will need a balanced equation with moles, so let's gather all the information in one place.
CH₃C₆H₄NH₂·HCl + (CH₃CO)₂O ⟶ CH₃C₆H₄NHCOCH₃ + junk
V/mL: 70.
c/mol·L⁻¹: 0.167
For simplicity in writing , let's call p-toluidine hydrochloride A and N-acetyl-<em>p</em>-toluidine B.
The equation is then
A + Ac₂O ⟶ B + junk
1. Moles of A
2. Moles of B
The molar ratio is 1 mol B:1 mol A
Moles of B = moles of A = 12 mmol = 0.012 mol
Answer:
4.4g
Explanation:
Mass of CaCO3 = 10g
Mass of CaO = 5.6g
Mass of CO2 =?
Mass of CaCO3 = Mass of CaO + Mass of CO2
Mass of CO2 = Mass of CaCO3 — Mass of CaO
Mass of CO2 = 10 — 5.6
Mass of CO2 = 4.4g
Answer:
Percentage yield = 85.2%
Explanation:
Given data:
Mass of Mg = 21.3 g
Actual yield of MgO = 30.2 g
Percentage yield = ?
Solution:
Chemical equation:
2Mg + O₂ → 2MgO
Number of moles of Mg = mass/molar mass
Number of moles of Mg = 21.3 g / 24.3 g/mol
Number of moles of Mg = 0.88 mol
Now we will compare the moles of MgO with Mg.
Mg : MgO
2 : 2
0.88 : 0.88
Mass of MgO:
Mass of MgO= moles × molar mass
Mass of MgO= 0.88 mol × 40.3g/mol
Mass of MgO = 35.46 g
Actual yield of MgO = 30.2 g
Percentage yield:
Percentage yield = Actual yield/theoretical yield × 100
Percentage yield = 30.2 g/ 35.46 g × 100
Percentage yield = 85.2%