Answer:
3.125 × 10¹³ Electrons
Explanation:
Data provided in the question:
Charge per unit length on rod = 0.00500 C/m
Length of the rod = 1 mm = 1 × 10⁻³ m
Therefore, the total charge on the rod
= Charge per unit length on rod × Length of the rod
= 0.00500 C/m × ( 1 × 10⁻³ m)
= 5 × 10⁻⁶ C
Thus,
Number of electrons removed
= total charge on the rod ÷ Charge of an electron
= 5 × 10⁻⁶ ÷ (1.6 × 10⁻¹⁹)
= 3.125 × 10¹³ Electrons
Answer:
-1
Explanation:
According to this question, the oxidation state/number of H and O in C2H4O is +1 and -2 respectively.
The oxidation state of carbon in the compound can be calculated thus:
Where;
x represents the oxidation number of C
C2H4O = 0 (net charge)
x(2) + 1(4) - 2 = 0
2x + 4 - 2 = 0
2x + 2 = 0
2x = -2
Divide both sides by 2
x = -1
The oxidation number of C in C2H4O is -1.
The correct answer that would best complete the given statement above would be the word MIXTURE. <span>A substance which contains various elements or compounds without bonding or in fixed proportions is a mixture. When a chemical union occurs, this is when a compound is being formed. Hope this answer helps.</span>
Density will be mass/volume= 0.2/500 = 0.0004
Answer:
2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓
Ksp = [2s]² . [s] → 4s³
Explanation:
Ag₂CrO₄ → 2Ag⁺ + CrO₄⁻²
Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.
2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓ Ksp
That's the expression for the precipitation equilibrium.
To determine the solubility product expression, we work with the Ksp
Ag₂CrO₄ (s) ⇄ 2Ag⁺ (aq) + CrO₄⁻² (aq) Ksp
2 s s
Look the stoichiometry is 1:2, between the salt and the silver.
Ksp = [2s]² . [s] → 4s³
