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allochka39001 [22]
2 years ago
13

Write a balanced overall reaction from these unbalanced half-reactions sn and ag

Chemistry
1 answer:
loris [4]2 years ago
8 0
Balance each one by adding electrons to make the charges on both sides the same:

Sn--> Sn2+ + 2 e-
Ag+ + 1 e- --> Ag

Now, you have to have the same number of electrons in the two half-reactions, so multiply the second one by 2 to get:

2 Ag+ + 2 e- --> 2 Ag

Now, just add the two half reactions together, cancelling anything that's the same on both sides:

2 Ag+ + Sn --> Sn2+ + 2 Ag

And you're done.
You might be interested in
Answer the questions in the table below about the shape of the phosphorus hexachloride (PCI^1_6) anion. What word or two-word ph
REY [17]

Shape of PCl₆⁻¹, is octahedral.

Hybrdisation of PCl₆⁻¹ is sp³d².

This can be seen the image attached.

The angle between phosphorous and chlorine is 90⁰C. As it has octahedral geometry.

So the answer is as follows:

Shape of PCl₆⁻¹, is octahedral.

Hybrdisation of PCl₆⁻¹ is sp³d².

The angle between phosphorous and chlorine is 90⁰C.

6 0
3 years ago
When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac
STatiana [176]

Answer:

\Delta _{comb}H=-2,093\frac{kJ}{mol}

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

Q_{rxn}+Q_{cal}=0

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\  \Delta _{comb}H=-2,093\frac{kJ}{mol}

Best regards!

4 0
2 years ago
How would you make a 2.00L of 0. 500M sodium chloride solution. (assume you have a fully equipped lab with water); sketch, calc
Gwar [14]

Answer:

Sodium chloride solution:

First you need to calculate the mass of salt needed (done in the explanation), which is 58.44g. Then it have to be weighted in an analytical balance in a weighting boat and then transferred into a 2L volumetric flask that is going to be filled until the mark with distilled water.

Sulfuric acid dilution:

First you need to calculate the volume needed (done in the explanation), it is 16.6 mL. Using a graduated pipette one measures this volume and transfer it into a 2L volumetric flask that is already half filled with distilled water, and then one fills it until its mark.

Explanation:

Sodium chloride solution:

Each liter of a 0.500M solution has half mol, so 2L of said solution has 1 mol of salt. Sodium chloride molar mass is 58.44g/mol, so in 2L of solution there is 58.44g of salt. That`s the mass that`s going to be weighted and transferred to a 2L volumetric flask.

Sulfuric acid dilution:

This is the equation for dilution of solutions:

c_{1} v_{1} =c_{2} v_{2}

Where "c1" stands for the initial concentration (stock solution concentration), "v1" for the initial volume (volume of stock solution used), "c2" for the desired concentration and "v2" for the desired volume.

When we are diluting from a stock solution we want to know how much do we have to pipette from the stock solution into our volumetric flask. We do so by isolating the "v1" term from the dilution equation:

v_{1} =\frac{c_{2} v_{2} }{c_{1} }

in this case that would be:

v_{1} =\frac{0.100 x2.0 }{12.0 }=0.0166L=16.6mL

5 0
3 years ago
A galvanic cell consists of a Cu(s)|Cu2+(aq) half-cell and a Cd(s)|Cd2+(aq) half-cell connected by a salt bridge. Oxidation occu
Kryger [21]

Answer:

Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)

Explanation:

A galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs.

The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

As it is given that cadmium acts as anode, it must be on the left hand side and copper must be on right hand side.

Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)

6 0
3 years ago
What vitamin is produced by a mammalian skin?
Contact [7]
It is Vitamin D, hope that helps
7 0
2 years ago
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