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denis-greek [22]
3 years ago
8

Consider the following reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) A reaction mixture is made containing an initial [SO2Cl2] of 2.2×10−2M

. At equilibrium, [Cl2]= 1.3×10−2M .
Chemistry
1 answer:
lukranit [14]3 years ago
5 0
I think you want to ask about Keq. At equilibrium, we can know [SO2Cl2] is 2.2*10-2 M -1.3*10-2M=9*10^-3 M. And [SO2]=[Cl2]. So the Keq=1.88*10^-2.
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Select all of the BENEFITS of a PARALLEL circuit. If one bulb burns out the rest of the bulbs will stay lit. The bulbs do not ge
artcher [175]

Answer:

The first two options are correct

Explanation:

The first two options are part of the benefits of a parallel connection of bulbs in a circuit. Here, the voltage of each connecting bulb is the same as the voltage of the bulb in the circuit hence all the bulbs have the same voltage running through them. Thus, when one bulb is removed/burns out, it does not affect the remaining bulbs (those ones will remain lit). Also, the addition of bulb(s) does not cause the remaining bulbs in the circuit to get dimmer (since they will all have the same voltage).

4 0
3 years ago
If a concentration of 10 fluorescent molecules per μm2of cell membrane is needed to visualize a cell under the microscope, how m
lara [203]

Answer : Total molecules that will be needed to visualize a single egg will be 78500 molecules of dye.

Explanation : As a single egg cell has an approximately diameter of 100 μm.

We can use this formula to calculate area of the cell membrane;

A = π (100)^{2} / 4;  

We can take π as 3.14 and we get;

A = 3.14 X (100)^{2} / 4  

Soving we get;

A =  7850 μm^{2}  

Here we have to calculate the amount of dye molecules which will be needed for 10 fluorescent molecules / μm^{2}  but;

here 1 μm^{2} = 7850 μm^{2} dye molecules.

Therefore, 10 fluorescent molecules will need;  

7850 X 10 = 78500 molecules of dye.

Therefore, the answer is 78500 molecules of dye.

6 0
3 years ago
Write the electron configuration (ground and excited states) for element with atomic number 16 ?
Talja [164]
2, 8,6 because it has to be in a configuration of 2,8,8
3 0
3 years ago
Which electron configuration represents a Scandium atom?
VARVARA [1.3K]
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hope this helped!
4 0
4 years ago
Calculate the concentrations of all species present in a 0.26 M solution of ethylammonium chloride (C2H5NH3Cl).
Alina [70]

Answer:

0.00000223

Explanation:

pKa for C2H5NH3+ = 10.7

pKw = 14.0

pKa + pKb = pKw

10.7 + pKb = 14.0

pKb = 14.0 - 10.7

pKb = 3.30

C2H5NH3Cl is a salt of ethylamine and HCl so it will dissolve in water to produce  C2H5NH3^+ + Cl^-

The base hydrolysis reaction:  C2H5NH3^+(aq) + H2O(l) <=> C2H5NH2(aq) + H3O^+(aq)

This reaction is described by Kb.

Kb = [C2H5NH2][H3O^+]/[C2H5NH3^+]

Let [C2H5NH2] = [H3O^+] = x,

so [C2H5NH3^+] = 0.26 - x

Kb = x^2/(0.26 - x) = 2.00 x 10^-11  

Let's solve for x. In this equation,  It is possible to solve without the use quadratic equation. So we can assume that 0.26 - x  is approximately equal to 0.26.  We won't know until we do the calculation.

We get:  x^2 + 2.00 x 10^-11x - 4.99 x 10^-12 = 0

With the use of a quadratic calculator.

x = 2.23 x 10^-6 M = [C2H5NH2] = [H3O^+]

0.26 - x  is just 0.26 M in this problem because 2.23 x 10^-6 M is insignificant.

[C2H5NH3^+] = 0.26 M = [Cl^-]

NOTE:

pH = -log [H3O^+] = -log(2.23 x 10^-6) = 5.65

Ka is the acid dissociation constant

Kb is the base dissociation constant

5 0
3 years ago
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