Answer:
well she can test both of the soap by putting one on and plate and another on the other plate and which ever is cleaner is your answer
Answer:
Explanation:
Assume we have 100g of this substance. That means we would have 20.24g of Cl and 79.76g of Al. Now we can find how many moles of each we have:
= 2.25 mol of chlorine
= 0.750 mol of Al.
To form a integer ratio, do 2.25/0.75 = 2.99999 ~= 3.
So the ratio is essentially Al : Cl => 1 : 3. To the compound is possibly 
.
However, it says it has a molar mass of 266.64 g/mol, and since AlCl3 has a molar mass of 133.32, it must be 
.
Actually this molecule isn't exactly AlCl3 (which is ionic). Al2Cl6 forms a banana bond where Cl acts as a hapto-2 ligand. But that's a bit advanced. All you need to know is X = Al2Cl6
She would observe a yellowish solid precipitate which is the lead iodide and a white solid precipitate which is the potassium nitrate.
This is because the lead nitrate solution which contains particles of lead will mix with the potassium iodide solution containing particles of iodide. Upon mixing,the lead particles from the Lead nitrate solution combines with the iodide particles from Potassium iodide and form two compounds, a yellowish solid precipitate called lead iodide and a white solid precipitate called Potassium nitrate.
The formation of entirely two new compounds is known as the double displacement reaction and can be written in a chemical equation as
2KI(aq.)+Pb(NO₃)₂(aq.)------>2KNO₃(aq.)+PbI ₂(s)
See similar answer here :https://brainly.in/question/46262462
The reaction between NaOH and H₂SO₄ is as follows;
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
NaOH is a strong acid and H₂SO₄ is a strong acid, therefore complete ionization into their respective ions takes place.
number of acid moles reacted - 0.112 M / 1000 mL/L x 39.1 mL = 0.0044 mol
the number of base moles required for neutralisation = 0.0044 x 2 = 0.0088 mol
Number of NaOH moles in 25.0 mL - 0.0088 mol
Therefore in 1000 mL - 0.0088 mol/ 25.0 mL x 1000 mL/L = 0.352 mol/L
Therefore molarity of NaOH - 0.352 M
