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tankabanditka [31]
3 years ago
9

What kind of waves do cellar telephones use to transmit amend recive signals

Chemistry
1 answer:
adelina 88 [10]3 years ago
4 0
The answer is microwave
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How can You friend electrons, protons and neutrons in an element?
Charra [1.4K]
You cant its not really that possible there diffrent elements to gather them into one source would be difficult more than likely a scientific explanation would answer this based upon test and research
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3 years ago
Muscle Activity
sasho [114]
I’m pretty sure it would be B
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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 the number of moles of gas is doubled at constant temperature and volume, the pressure of the gas:__________
Vladimir79 [104]

Answer:

answer is b

use pv=nRT

p directly proportional to moles(n)

p1/p2=n1/n2

p/p2=n/2n

p2=2p

4 0
4 years ago
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krek1111 [17]

Answer:

Dubnium: Group 5 Period 7

Hydrogen: Group 1 Period 1

Explanation:

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