Answer:
give the ion X
<h2>X2+</h2>
Explanation:
it is positively charged because the protons are more than electrons
Answer:
The answer to your question is letter b) X
Explanation:
Data
W = 5 units
X = 6 units
Chemical reaction
2W + 3X ⇒ 3Y + Z
To determine which reactant is the limiting reactant, we must use proportions:
-Theoretical proportion W / X = 2 / 3 = 0.67
- Experimental proportion W / X = 5 / 6 = 0.83
As the experimental proportion is higher than the theoretical proportion, we conclude that the amount of W is higher in the experiment so the limiting reactant is X.
Answer:
Water has the greatest ΔEN
ΔEN H₂O → 3.4 - 2.1 = 1.3 Option D.
Explanation:
We should find the Electronegativity data in the Periodic table for all the elements:
C : 2.6
O: 3.4
H: 2.1
S: 2.6
N: 3.0
a. ΔEN CO₂ → 3.4 - 2.6 = 0.4
b. ΔEN H₂S → 2.6 - 2.1 = 0.5
c. ΔEN NH₃ → 3 - 2.1= 0.9
d. ΔEN H₂O → 3.4 - 2.1 = 1.3
Q1)
NaOH solution is in the burette and called the titrant
HCl a known volume is in the titration flask and called the titrand.
initial burette reading of NaOH - 0.33 mL
final buretter reading of NaOH - 24.19 mL
therefore volume of NaOH added / dispensed - 24.19 - 0.33 = 23.86 mL
23.86 mL of NaOH was dispensed
Q2)
phenolphthalein is an acid base reaction indicator that shows a colour change from colourless in neutral and acidic media to pink colour in basic media.
phenolphthalein when added to the acid solution is colourless.
Once NaOH is added, OH⁻ reacts with H⁺ in acid and is neutralised. After all the H⁺ in medium is used up, the added OH⁻ is no longer neutralised.
in the presence of excess OH⁻, solution becomes basic and phenolphthalein becomes pink.
number of moles of NaOH = concentration x volume
number of NaOH moles = 0.1550 mol/L x (23.86 x 10⁻³ L) = 3.698 x 10⁻³ mol
NaOH moles added = 3.698 x 10⁻³ mol
Q3)
Neutralisation reaction is the reaction between an acid and base.
H⁺ ions of acid and OH⁻ ions of base react to form water and become neutralised.
when acid and base react they form water and salt.
the reaction is as follows;
HCl (aq) + NaOH (aq) --> NaCl (aq) + H₂O (l)
