Answer:
Explanation:
» The prediction is 98% correct because single displacement reaction type is highly possible.
This is because Fluorine has is more electronegative than Chlorine in Potassium Chloride. So, it strongly displaces Chlorine from the solution hence forming Chlorine gas.
» The 2% of wrong prediction maybe because of wrong reactant measurements following mole concept chemistry.
If you are asked the observation,
Observation » <u> </u><u>A</u><u> </u><u>green</u><u> </u><u>yellowish</u><u> </u><u>gas</u><u> </u><u>is</u><u> </u><u>formed</u><u>.</u>
This gas is Chlorine gas (Cl2)
Answer:
H2Br + 2KOH ----- K2Br + 2H2O
Explanation:
Depression in Freezing point
= Kf × i × m
where m is molality , i is Van't Hoff factor, m = molality
Since molality and Kf remain the same
depression in freezing point is proportional to i
i= 2 for CuSO4 ( CuSO4----------> Cu+2 + SO4-2
i=1 for C2h6O
i= 3 for MgCl2 ( MgCl2--------> Mg+2+ 2Cl-)
So the freezing point depression is highest for MgCl2 and lowest for C2H6O
so freezing point of the solution = freezing point of pure solvent- freezing point depression
since MgCl2 has got highest freezing point depression it will have loweest freezing point and C2H6O will have highest freezing point
<span>The mass of an atom located in the A) nucleus.</span>
Answer:
- The answer is the concentration of an NaOH = 1.6 M
Explanation:
The most common way to solve this kind of problem is to use the formula
In your problem,
For NaOH
C₁ =?? v₁= 78.0 mL = 0.078 L
For H₂SO₄
C₁ =1.25 M v₁= 50.0 mL = 0.05 L
but you must note that for the reaction of NaOH with H₂SO₄
2 mol of NaOH raect with 1 mol H₂SO₄
So, by applying in above formula
- (C₁ * 0.078 L) = (2* 1.25 M * 0.05 L)
- C₁ = (2* 1.25 M * 0.05 L) / (0.078 L) = 1.6 M
<u>So, the answer is the concentration of an NaOH = 1.6 M</u>
