→ 
Explanation:
- The products formed are chromic chloride and cobalt.
Chromium + Cobaltous Chloride = Chromic Chloride + Cobalt
- Type of reaction is Single Displacement (Substitution) which is there is a displacement of one atom.
Reactants used in the reaction are -
- Chromium
- Cobaltous Chloride
Products formed in the reaction are -
- Chromic Chloride
- Cobalt
Hence, the chemical reaction is as follows -
→
For balancing the above chemical equation we need to add a coefficient of 2 in front of chromium and of 3 in front of cobalt(II)chloride on right-hand-side while of 2 in front of chromium chloride and of 3 in front of carbon monoxide on left-hand-side of the equation.
Hence, the balanced equation is -
→
Answer:
An ion is defined as an atom or molecule that has gained or lost one or more of its valence electrons, giving it a net positive or negative electrical charge.
Explanation:
I am here zinda -_+...
Answer: -
The first step involves protonation of the carbonyl oxygen.
After protonation, the Alcohol oxygen now attacks the carbon of the carbonyl.
Thus a six membered ring is formed with 5 carbon atoms and 1 oxygen atom. The 1st position carbon atom has 2 OH groups.
One of these gets again protonated.
This leaves as water. With the loss of the H+, there results a carbonyl at 1 position.
Thus 5-hydroxypentanoic acid forms a lactone or 2-oxanone in presence of acid.
First, you need to count copper mass in alloy.
Second, you have to make an equation an find x ( the copper mass must be added). The answer is: 13,5g pure copper
Answer:
1.7 × 10 ^42
Explanation:
Using Nernst equation
E°cell = RT/nF Inq
at equilibrium
Q=K
E°cell = 0.0257 /n Ink= 0.0592/n log K
Fe2+(aq)+2e−→Fe(s) E∘= −0.45 V
Ag+aq)+e−→Ag(s) E∘= 0.80 V
Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s)
balance the reaction
Fe → Fe²⁺ + 2e⁻ reversing for oxidation E° = 0.45 v
2 Ag⁺ +2e⁻ → 2Ag
n = 2 moles and K = equilibrium constant
E° cell = 0.80 + 0.45 = 1.25 V
E° cell = (0.0592 / n) log K
substitute the value into the equations and solve for K
(1.25 × 2) / 0.0592 = log K
42.23 = log K
k = 10^ 42.23
K = 1.7 × 10 ^42
