Answer: This contains magnesium, Mg2+, and hydroxide, OH–
, ions. Each magnesium ion is +2 and
each hydroxide ion is -1: two -1 ions are needed for one +2 ion, and the formula for magnesium
hydroxide is Mg(OH)2. The (OH)2 indicates there are two OH–
ions. In a formula unit of
Mg(OH)2, there are one magnesium ion and two hydroxide ions; or one magnesium, two
oxygen, and two hydrogen atoms. The subscript multiplies everything in ( )
hope that helped!!
The balanced chemical reaction is written as:
<span>4C(s) + S8(s) → 4CS2(l)
We are given the amount of carbon and sulfur to be used in the reaction. We need to determine first the limiting reactant to be able to solve this correctly.
</span>7.70 g C ( 1 mol / 12.01 g) =0.64 mol C
19.7 g S8 ( 1 mol / 256.48 g) = 0.08 mol S8
The limiting reactant would be S8. We use this amount to calculate.
0.08 mol S8 ( 4 mol CS2 / 1 mol S8 ) ( 256.48 g / 1 mol ) = 78.8 g CS2
<span>A solution with a pH of 4 has ten times the concentration of H</span>⁺<span> present compared to a solution with a pH of 5.
</span>pH <span>is a numeric scale for the acidity or basicity of an aqueous solution. It is the negative of the base 10 logarithm of the molar concentration of hydrogen ions.
</span>[H⁺] = 10∧-pH.
pH = 4 → [H⁺]₁ = 10⁻⁴ M = 0,0001 M.
pH = 5 → [H⁺]₂ = 10⁻⁵ M = 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 0,0001 M / 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 10.
Zn, Cd, and Ag are transition metals that usually form only one monoatomic cation.
A monatomic cation is a cation made of only one atom.
Cations are positively charged ions, in this example Ag⁺, Cd²⁺ and Zn²⁺.
These cations form only one type of ion, while iron and copper form more than one type of cations.
Iron and copper form cations with different charges (Fe²⁺, Fe³⁺, Cu⁺, Cu²⁺).
It depends on electron configuration which type would be formed.
Electron configuration of zinc atom: ₃₀Zn 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²
Transition metals are elements in the d-block of the Periodic table.
More about transition metals: brainly.com/question/12843347
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Answer:
-1.05 V
Explanation:
A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.
Oxidation half equation;
Sn(s) ------> Sn^2+(aq) + 2e
Reduction half equation:
Mn^2+(aq) + 2e ----> Mn(s)
Cell voltage= E°cathode - E°anode
E°cathode= -1.19V
E°anode= -0.14 V
Cell voltage= -1.19 V - (-0.14V)
Cell voltage= -1.05 V
