<h2>
Hey There!</h2><h2>
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Answer:</h2><h2>
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ELECTROPLATING </h2>
Electroplating the plating one metal on to the another metal, It is mostly used for preventing corrosion by using copper or chromium or decorate the object by using gold or silver plating.
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<h3>Principle:</h3>
When electricity is passed the thin layer of metal is deposited on another metal and water molecule given out as a By-Product, Thus this process works on the principle of Hydrolysis.
<h2>_____________________________________</h2><h2>QUESTION:</h2>
A) Electrolysis
B) Chromium prevents corrosion and gives the fine shining touch to the objects.
C) The metal which is deposited to the object i.e. spoon will be connected to the positive electrode of a battery, Thus it is anode. The spoon at which electroplating is need to be done is connected to the negative electrode, thus the Spoon is cathode.
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Best Regards,</h2><h2>
'Borz'</h2>
Answer:
The rain falling in New England is 2.29 times more acidic than the one in the American Midwest.
Explanation:
The acidity of a solution depends on the concentration of H⁺ ions ([H⁺]). We can calculate this concentration from the pH using the following expression.
pH = -log ([H⁺])
American Midwest
pH = -log ([H⁺])
5.02 = -log ([H⁺])
[H⁺] = antilog (-5.02) = 9.55 × 10⁻⁶ M
New England
pH = -log ([H⁺])
4.66 = -log ([H⁺])
[H⁺] = antilog (-4.66) = 2.19 × 10⁻⁵ M
The ratio of concentrations is:
The rain falling in New England is 2.29 times more acidic than the one in the American Midwest.
Answer:
Single Displacement reaction
In a displacement reaction, a more reactive element replaces a less reactive element from a compound.
Change in colour takes place with no precipitate forms.
Metals react with the salt solution of another metal.
Examples:
2KI + Cl2 → 2KCl + I2
CuSO4 + Zn → ZnSO4 + Cu
Double displacement reaction
In a double displacement reaction, two atoms or a group of atoms switch places to form new compounds.
Precipitate is formed.
Salt solutions of two different metals react with each other.
Examples:
Na2SO4 + BaCl2 → BaSO4 + 2NaCl
2KBr + BaCl2 → 2KCl + BaBr2
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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>
Na 1s²2s²2p⁶3s¹
↓ - e⁻
Na⁺ 1s²2s²2p⁶ 2+2+6=10 e⁻
10 electrons are in sodium ion Na⁺
