Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
There are 0.0214 mol Cs in 1.29×10²² atoms Cs.
Use <em>Avogadro’s number</em> to convert atoms of Cs to moles of Cs.
Moles of Cs = 1.29×10²² atoms Cs × (1 mol Cs /6.022 × 10²³ atoms Cs) = 0.0214 mol Cs
Answer:
Yes. The volume would be 1/4 of the initial volume.
Explanation:
At constant temperature, the pressure of a gas is inversely proportional to the volume of the gas. Hence;
P1V1 = P2V2
<em>where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.</em>
If the pressure of a gas is quadrupled;
P2 = 4P1, the equation becomes
P1V1 = 4P1 x V2
Making V2 the subject:
V2 = P1V1/4P1
V2 = V1/4
<em>This means that the volume would change by being reduced to </em><em>1/4 </em><em>of the initial volume.</em>
Convert all the gases from grams to moles using their molar masses first. Remember that nitrogen and oxygen exist as DIATOMIC gases.
O2 - 25 g / 32 g/mol = 0.78 mol O2
N2 - 15 g / 28.02 g/mol = 0.54 mol N2
He - 10 g / 4.0 g/mol = 2.5 mol He
Add all the moles of gases.
0.78 + 0.54 + 2.5 = 3.82 moles
Divide O2 moles by total moles.
0.78/3.82 = 0.20
As is, the equation has:
Reactants:
2 K
2 O
2 H
Products:
1 K
1 O
1 H
So, we can add coefficients to balance the equation:
K2O + H2O = 2KOH
The coefficient means that there are 2 moles of KOH, and now the equation is balanced as there is 2 of everything on both sides.
