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vova2212 [387]
3 years ago
15

How does the size of ions affect the conductivity of a solution?

Chemistry
1 answer:
mrs_skeptik [129]3 years ago
4 0

Small ions have small areas. There is less resistance as they move through the solution.

For example, in molten salts, the conductivity of <span>Li+</span> is greater than that of <span>Cs+</span>.

Small ions have high charge density.

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If the temperature of an area drops 8 degrees from one day the next the climate has changed.
vlabodo [156]
I would say false hope that helped
3 0
3 years ago
The process of fermentation involves chemical reactions that convert solid glucose (C6H12O6) into aqueous ethanol and carbon dio
e-lub [12.9K]

Answer:

C6H12O6 —> 2C2H5OH + 2CO2

Explanation:

The equation for the reaction is given below:

C6H12O6 —> C2H5OH + CO2

We can balance the equation above as follow:

There are 12 atoms of H on the left side and 6 atoms of the right side. It can be balance by putting 2 in front of C2H5OH as shown below:

C6H12O6 —> 2C2H5OH + CO2

There are 6 atoms of C on the left side and 5 atoms on the right side. It can be balance by putting 2 in front of CO2 as shown below:

C6H12O6 —> 2C2H5OH + 2CO2

Now the equation is balanced.

4 0
3 years ago
What will happen if a peeled banana is put on a hotplate?
inysia [295]
It will melt or the molecules inside of it will get hot
8 0
3 years ago
1. If I have 45 L of He in a balloon at 25 degrees celsius and increase the temperature of the
Greeley [361]

Use Charles' Law: V1/T1 = V2/T2. We assume the pressure and mass of the helium is constant. The units for temperature must be in Kelvin to use this equation (x °C = x + 273.15 K).

We want to solve for the new volume after the temperature is increased from 25 °C (298.15 K) to 55 °C (328.15 K). Since the volume and temperature of a gas at a constant pressure are directly proportional to each other, we should expect the new volume of the balloon to be greater than the initial 45 L.

Rearranging Charles' Law to solve for V2, we get V2 = V1T2/T1.  

(45 L)(328.15 K)/(298.15 K) = 49.5 ≈ 50 L (if we're considering sig figs).

7 0
2 years ago
Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /
barxatty [35]

Answer:

[Pb^{2+}]=3.9 \times 10^{-2}M

this is the concentration required to initiate precipitation

Explanation:

PbCl_2  ⇄ Pb^{2+}+2Cl^-

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

Ip=[Pb^{2}][2Cl^-]^2=Ksp

Ksp=2.4\times 10^{-4}

lets solubility=S

[Pb^{2+}] = S

[Cl^-]=2S

Ksp=[Pb^{2+}]\times [Cl^-]^2

Ksp=S \times (2S)^2

Ksp=4S^3

S=\sqrt[3]{\frac{Ksp}{4} }

S=3.9\times 10^{-2}

[Pb^{2+}]=3.9 \times 10^{-2}M this is the concentration required to initiate precipitation

4 0
3 years ago
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