Half-reaction for the cell's anode is given below:
Anode : 
The anode is defined as the electrode at which electrons leave the cell and oxidation occurs, and the cathode as the electrode at which electrons enter the cell and reduction occurs. The anode is usually the positive side.
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Your given question is quite incomplete here is complete question.
A voltaic cell is based on the reduction of _ Agt(aq) to Ag(s) and the oxidation of Sn(s) to Sn2+(aql) : Part 1 Include the phases of all species in the chemical equation: (aqh Anode: Sn(s) Sn?+ (aq)
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Answer:
Aluminium (Al): (3*2)+(5*2)=16
Sulphor (S): (3*1)=3
Oxygen (O): (4*3)+(3*1)=15
The answer for the following problem is mentioned below.
- <u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.</em></u>
Explanation:
Given:
mass of calcium phosphate (
) = 125.3 grams
We know;
molar mass of calcium phosphate ( ) = (40×3) + 3 (31 +(4×16))
molar mass of calcium phosphate ( ) = 120 + 3(95)
molar mass of calcium phosphate ( ) = 120 +285 = 405 grams
<em>We also know;</em>
No of molecules at STP conditions(
) = 6.023 × 10^23 molecules
To solve:
no of molecules present in the sample(N)
We know;
N÷
=
N =(405×6.023 × 10^23) ÷ 125.3
N = 19.3 × 10^23 molecules
<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules</em></u>
Answer:
1.2×10²³ atoms.
Explanation:
Data obtained from the question include:
Mole of propanone = 0.20 mole
Number of atoms of propanone =.?
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ atoms.
This implies that 1 mole of propanone also contains 6.022×10²³ atoms.
Thus, we can obtain the number of atoms in 0.20 mole of propanone as illustrated below:
1 mole of propanone contains 6.022×10²³ atoms.
Therefore, 0.20 mole of propanone will contain = 0.2 × 6.022×10²³ = 1.2×10²³ atoms.
Thus, 0.20 mole of propanone contain
1.2×10²³ atoms.
<span>Determine the root-mean-square sped of CO2 molecules that have an average Kinetic Energy of 4.21x10^-21 J per molecule. Write your answer to 3 sig figs.
</span><span>
E = 1/2 m v^2
If you substitute into this formula, you will get out the root-mean-square speed.
If energy is Joules, the mass should be in kg, and the speed will be in m/s.
1 mol of CO2 is 44.0 g, or 4.40 x 10^1 g or 4.40 x 10^-2 kg.
If you divide this by Avagadro's constant, you will get the average mass of a CO2 molecule.
4.40 x 10^-2 kg / 6.02 x 10^23 = 7.31 x 10^-26 kg
So, if E = 1/2 mv^2
</span>v^2 = 2E/m = 2 (4.21x10^-21 J)/7.31 x 10^-26 kg = 115184.68
Take the square root of that, and you get the answer 339 m/s.
