
Here's the balanced equation for given Double displacement reaction ~

The products fored are : Lead Iodide ( PbI2 ) and Potassium Nitrate ( KNO3 )
Answer:
B
Explanation:
The tendency of a liquid to convert to vapour increases smoothly with increasing temperature. Vapour pressure shows the tendency of a liquid to convert to vapour. Increase In vapour pressure shows an increased tendency to convert to vapour. The higher the temperature, the higher the vapour pressure.
Strong electrolytes are completely ionised in solution.
Weak ones are only partially ionised.
Non electrolytes do not form ions
Answer:
the concentration of the solution is 0.00906 M
Explanation:
Given the data in the question;
we know that from Nernst Equation;
E = E⁰ - ((0.0592/n) logQ)
now, E₀ for concentration cell is 0
n for this redox is 2
concentration of the unknown solution is x
so we substitute
0.045 = 0 - ( 0.0592 / 2)log( x/0.300 ))
0.045 = -0.0296log( x/0.300 )
divide both side by 0.0296
1.52 = -log( x/0.300 )
x/0.300 =
x/0.300 = 0.0301995
we cross multiply
x = 0.300 × 0.0301995
x = 0.00906 M
Therefore, the concentration of the solution is 0.00906 M
Answer:
(a) I⁻ (charge 1-)
(b) Sr²⁺ (charge 2+)
(c) K⁺ (charge 1+)
(d) N³⁻ (charge 3-)
(e) S²⁻ (charge 2-)
(f) In³⁺ (charge 3+)
Explanation:
To predict the charge on a monoatomic ion we need to consider the octet rule: atoms will gain, lose or share electrons to complete their valence shell with 8 electrons.
(a) |
I has 7 valence electrons so it gains 1 electron to form I⁻ (charge 1-).
(b) Sr
Sr has 2 valence electrons so it loses 2 electrons to form Sr²⁺ (charge 2+).
(c) K
K has 1 valence electron so it loses 1 electron to form K⁺ (charge 1+).
(d) N
N has 5 valence electrons so it gains 3 electrons to form N³⁻ (charge 3-).
(e) S
S has 6 valence electrons so it gains 2 electrons to form S²⁻ (charge 2-).
(f) In
In has 3 valence electrons so it loses 3 electrons to form In³⁺ (charge 3+).