Answer: pH of HCl =5, HNO3 = 1,
NaOH = 9, KOH = 12
Explanation:
pH = -log [H+ ]
1. 1.0 x 10^-5 M HCl
pH = - log (1.0 x 10^-5)
= 5 - log 1 = 5
2. 0.1 M HNO3
pH = - log (1.0 x 10 ^ -1)
pH = 1 - log 1 = 1
3. 1.0 x 10^-5 NaOH
pOH = - log (1.0 x 10^-5)
pOH = 5 - log 1 = 5
pH + pOH = 14
Therefore , pH = 14 - 5 = 9
4. 0.01 M KOH
pOH = - log ( 1.0 x 10^ -2)
= 2 - log 1 = 2
pH + pOH = 14
Therefore, pH = 14 - 2 = 12
The average acceleration is -5.0 m·s⁻².
The formula for acceleration (<em>a</em>) is
= 25 m·s⁻¹; 
= 0; 
= 5.0 s
∴ 
= -5.0 m·s⁻²
The negative sign tells you that the object is <em>slowing down</em>, i.e., it is <em>decelerating</em>.
Answer:
No. of atom =
no.of moles x avagardro's number xatomicity
= weight /molar mass x No x atomicity
=8.2/142 x6.02x10^23 x 4
=0.346 x 10^23(approximately)
Answer is: n<span>o, because the ion product is less than the Ksp of lead iodide. </span>
Chemical dissociation 1: KI(s) → K⁺(aq) + I⁻(aq).
Chemical dissociation 2: Pb(NO₃)₂(s) → Pb²⁺(aq) + 2NO₃⁻(aq).
Chemical reaction: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s).
Ksp(PbI₂) = 7.1·10⁻⁹.
V = 500 mL ÷ 1000 mL/L = 0.5 L.
c(KI) = c(I⁻) = 0.0025 mol ÷ 0.5 L.
c(I⁻) = 0.005 M.
c(Pb(NO₃)₂) = c(Pb²⁺) = 0.00004 mol ÷ 0.5 L.
c(Pb²⁺) = 0.00008 M.
Q = c(Pb²⁺) · c(I⁻)².
Q = 8·10⁻⁵ M · (5·10⁻³ M)².
Q = 2·10⁻⁹; <span> the ion product.</span>
Answer:
D) Sulphur
Explanation:
Sulphur has the strongest electromagnetivity here
