Answer:
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You are given a galvanic cell consists of a Ni²⁺/ Ni half-cell and a standard hydrogen electrode. Also, you are given that the half cell Ni²⁺/ Ni will act as an anode, and the standard cell potential is 0.26V. You are asked to find the standard reduction potential for the half cell Ni²⁺/ Ni.
You will have a half - reaction for both nickel and hydrogen
The conversion of the symbol Ni²⁺/ Ni half-cell is
Ni²⁺ + 2e⁻ → Ni (s) E = 0.26V
and the conversion of the standard hydrogen electrode (SHE) is
2H⁺ + 2e⁻ → H₂ (g) E = 0V
Since H⁺ ions is a it difficult to set up during the process, nickel will be deposited at the cathode side instead of the anode. Therefore, The standard electron potential of the nickel will have -0.26V.
Answer:
it is metals, it is the most prominent type of element on the table
Answer:
Explanation: See images below for explanation
Answer:
[Cl⁻] = 1,5x10⁻⁴M
Explanation:
First of all, let's determinate the mole of each salt.
Molarity . volume = Mole
Volume must be in L, cause molarity is mol/L
NaCl → Na⁺ + Cl⁻
Ratio is 1:1
0.15 mol/L . 0.025L = 3.75x10⁻³ mole
As ratio is 1:1, from 3.75x10⁻³ mole of salt, I have 3.75x10⁻³ mole of chloride
CaCl₂ → Ca²⁺ + 2Cl⁻
Ratio is 1:2 so, from 1 mol of salt I'll get the double of mole of chloride
0.075 mol/L . 0.010 L = 7.5x10⁻⁴ mol
7.5x10⁻⁴ mol . 2 = 1.5x10⁻³ mole
Total mole of Cl⁻: 3.75x10⁻³ + 1.5x10⁻³ = 5.25x10⁻³
This 5.25x10⁻³ mole are present in a total volume of 35 mL.
Let's convert 35 mL in L → 0.035L (35/1000)
Molarity is mol/L → 5.25x10⁻³ mol / 0.035L = 1,5x10⁻⁴M