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bearhunter [10]
3 years ago
14

A certain metal M forms a soluble nitrate salt M(NO3), Suppose the left half cell ofa galvanic cell apparatus is filled with a 4

.00 M solution of M (NO,), and the right half cell with a 20.0 mM solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 35.0 °C.
Required:
a. Which electrode will be positive?
b. What voltage will the voltmeter show?
Chemistry
1 answer:
kozerog [31]3 years ago
6 0

Answer:

Explanation:

The left electrode will be positive on the grounds that focus on the concentration that the cell electron moves from a lower concentration fixation to a higher concentration. Thus right electrode will go about and act as an anode and will be negative.  Also, the left electrode will be the cathode and will be positive.

The concentration cell E_{Cell} = \dfrac{-0.0591}{1} log \dfrac{0.02}{4}

E_{Cell} =0.136 V

         = 136 mV

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An exhaled air bubble underwater at 290.
Vinil7 [7]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>

Explanation:

Given:

Initial pressure (P_{1}) = 290 kPa

Final pressure (P_{2}) = 104 kPa

Initial volume (V_{1}) = 18.9 ml

To find:

Final volume (V_{2})

We know;

From the ideal gas equation;

    P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of gas

n represents the no of the moles

R represents the universal gas constant

T represents the temperature of the gas

So;

   P × V = constant

   P ∝ \frac{1}{V}

From the above equation;

              \frac{P_{1} }{P_{2} }  = \frac{V_{2} }{V_{1} }

P_{1} represents the initial pressure of the gas

P_{2} represents the final pressure of the gas

V_{1} represents the initial volume of the gas

V_{2} represents the final volume of the gas

Substituting the values of the above equation;

                    \frac{290}{104} = \frac{V_{2} }{18.9}

             V_{2} = 52.7 ml

<u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>

6 0
3 years ago
My science exam is due tomorrow and I HAVE LIMITED TIME. can someone help me how to absorb anything you read and memorize it...
igor_vitrenko [27]

Answer:  

It won't let me type this for some reason but here it is.

7 0
2 years ago
Which ion in the ground state has the same electron configuration as an atom of neon in the ground state?
Ainat [17]
  • Cl-

Lets see how

  • Z for Cl is 9
  • One electron is added so new Z=10

Electronic configuration

  • 1s²2s²2p⁶

Or

  • [Ne]

option B is correct

6 0
2 years ago
Read 2 more answers
2.5 moles of a gas is enclosed in a 87.2 L cylinder with a moveable piston at 425 K and 1.0 atm. An additional 2.5 moles of gas
Nesterboy [21]

Answer:

112 L

Explanation:

Since the pressure is being held constant, you can use the following variation of the Ideal Gas Law to find the new volume:

\frac{V_1}{T_1N_1}=\frac{V_2}{T_2N_2}

In this equation, "V₁", "T₁", and "N₁" represent the initial volume, temperature, and moles. "V₂", "T₂", and "N₂" represent the final volume, temperature, and moles.

V₁ = 87.2 L                             V₂ = ? L

T₁ = 425 K                              T₂ = 273 K

N₁ = 2.5 moles                       N₂ = 2.5 + 2.5 = 5.0 moles

\frac{V_1}{T_1N_1}=\frac{V_2}{T_2N_2}                                                   <----- Formula

\frac{87.2 L}{(425K)(2.5 moles)}=\frac{V_2}{(273 K)(5.0 moles)}                    <----- Insert values

\frac{87.2 L}{1062.5}=\frac{V_2}{1365}                                                    <----- Simplify denominators

0.08207=\frac{V_2}{1365}                                                 <----- Simplify left side

112L={V_2}                                                        <----- Multiply both sides by 1365

4 0
2 years ago
How many atoms are in a sample of 68.7 grams copper (Cu)?
kondaur [170]
To find the number of atoms in the sample of copper, we need its atomic mass and the use of the Avogadro's number to find the equivalent units in terms of atoms.

68.7 g Cu ( 1 mol / 63.55 g ) ( 6.022 x 10^23 atoms / 1 mol ) = 6.51 x 10^23 atoms Cu
7 0
3 years ago
Read 2 more answers
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