<h3><u>Answer;</u></h3>
a) It allows electrons to flow from the anode to the cathode.
<h3><u>Explanation</u>;</h3>
- <em><u>Voltaic cell is an electrochemical cell in which a spontaneous chemical reaction produces the flow of electrons</u></em>.
- Electrons are produced by the oxidation reaction occurring at the anode. Electrons flow through the conducting wire from the anode to the cathode. At the cathode these electrons are used to reduce copper(II) ions to copper atoms.
- <em><u>A conducting wire or a wire play connects the two electrodes allowing electrons to flow from the anode to the cathode</u></em>.
Clutch Prep
Ch.2 - Atoms & ElementsSee all chapters
Atomic Theory
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Solution: Which of the following chemical reactions is/are NOT possible according to Dalton's atomic theory?a. reaction 1: CCl4 → CH4b. reaction 2: N2 + 3H2 → 2NH3c. reaction 3: 2H2 + O2 → 2H2O + Au
Problem
Which of the following chemical reactions is/are NOT possible according to Dalton's atomic theory?
a. reaction 1: CCl4 → CH4
b. reaction 2: N2 + 3H2 → 2NH3
c. reaction 3: 2H2 + O2 → 2H2O + Au
Part 1 :- Super gaint star have mass from 10 to 70 solar masses and brightness from 30,000 upto hundreds of thousand times the solar luminosity
blue super giant surface temperature is 20,000 to 50,000 degree Celsius example :0 Rigel its mass is around 20 times the sun mass it give light which 60,000 sun together give .
part 2 :- HR diagram : hertzsprung -Russel diagram ( attached below)
Answer:
[Ag⁺] = 5.0x10⁻¹⁴M
Explanation:
The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:
Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸
Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷
The PbI₂ <em>just begin to precipitate when the product [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>
<em />
As the initial [Pb²⁺] = 0.0050M:
[Pb²⁺] [I⁻]² = 1.4x10⁻⁸
[0.0050] [I⁻]² = 1.4x10⁻⁸
[I⁻]² = 1.4x10⁻⁸ / 0.0050
[I⁻]² = 2.8x10⁻⁶
<h3>[I⁻] = 1.67x10⁻³</h3><h3 />
So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:
[Ag⁺] [I⁻] = 8.3x10⁻¹⁷
[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷
<h3>[Ag⁺] = 5.0x10⁻¹⁴M</h3>
Answer:
fe3o4+4h2 - 3fe + 4h2o
therefore coefficient is 4
