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4 years ago

I am stuck on this lab and have absolutely no idea how to do it.

Chemistry

2 answers:

expeople1 [14]4 years ago

4 0

Hey i did this lab a few months ago!

seropon [69]4 years ago

4 0

Id get the elements chart that might help

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The ksp of pbbr2 is 6.60× 10–6. what is the molar solubility of pbbr2 in 0.500 m kbr solution

Alex_Xolod [135]

The solubility of PbBr₂(s) with the presence of 0.500 M of KBr is 2.64 x 10⁻⁵ M.

4 0

3 years ago

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For the reaction represented by the equation 2H2 + O2→ 2H2O, how many grams of water can be produced from 6.0 grams of O2?

leonid [27]

Answer:

a. 6 grams

b. 108 grams

c. 2 grams

d. 54 grams

5 0

3 years ago

A solution is prepared by diluting 50.00 ml of 2.575 m solution of hno3 to 250.0 ml. what is the molarity of the resulting solut

larisa [96]

Thank you for posting your question here at brainly. Below are the choices that can be found elsewhere:

12.88 M
0.1278 M 
0.2000 M 
0.5150 M

Below is the answer:

5 times diluted (250/50),so 2.575/5=0.515 M

I hope it helps.

8 0

3 years ago

Read 2 more answers

Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.85×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.17×10−4, what is the

kramer

Answer:

The equilibrium constant of the given reaction is 0.01351.

Explanation:

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

The equilibrium constant of the reaction = $K_3=1.85\times 10^{-10}$

$K_3=\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}$ ...[1]

$AgCl(aq)\rightleftharpoons Ag^{+}(aq)+Cl^-(aq)$

The equilibrium constant of the reaction = $K_4=1.17\times 10^{-4}$

$K_4=\frac{[Ag^+][Cl^-]}{[AgCl]}$ ..[2]

$[Cl^-]=\frac{K_4\times [AgCl]}{[Ag^+]}$

$PbCl_2(aq)+2Ag^+(aq)\rightleftharpoons 2AgCl(aq)+Pb^{2+}(aq)$

The expression of equilibrium constant of the creation is ;

$K=\frac{[AgCl]^2[Pb^{2}]}{[PbCl_2]][Ag^+]^2}$

Dividing [1] by [2]

$\frac{K_3}{K_4}=\frac{\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}}{\frac{[Ag^+][Cl^-]}{[AgCl]}}$

$\frac{K_3}{K_4}=\frac{[Pb^{2+}][Cl^-][AgCl]}{[PbCl_2][Ag^+]}$

Substituting the value of $[Cl^-]$ from [2] :

$\frac{K_3}{K_4}=\frac{[Pb^{2+}][AgCl]}{[PbCl_2][Ag^+]}\times \frac{K_4\times [AgCl]}{[Ag^+]}$

$\frac{K_3}{K_4}=K_4\times K$

$K=\frac{K_3}{(K_4)^2}=\frac{1.85\times 10^{-10}}{(1.17\times 10^{-4})^2}$

$K=0.01351$

The equilibrium constant of the given reaction is 0.01351.

3 0

3 years ago

If you were to reach the location in 316 minutes, what is your average speed in

nirvana33 [79]

The average speed is zero if our frame of reference is the rotation of the Earth because the relative distance with respect to the earth is zero.

Average speed = 0.26 km/hr

total time = 316 minutes

distance = speed * time

= 8.3 * 5*1/30 = 8.3 *1/6

= 1.38 km

Average speed = total distance / total time

1.38 km / 5.27

= 0.26 km/hr

Distance is described to be the importance or length of displacement among positions. observe that the gap between two positions is not the same as the distance traveled between them. Distance traveled is the whole period of the path traveled among positions. Distance traveled isn't a vector.

There are three foremost styles of average: imply, median, and mode. each of these strategies works barely otherwise and frequently results in slightly distinct ordinary values. The suggest is the maximum usually used commonly. To get the mean cost, you add up all of the values and divide this general by means of the variety of values.

Learn more about average speed here:-brainly.com/question/4931057

#SPJ4

4 0

2 years ago