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Sauron [17]
3 years ago
7

How does width affect capillary rise?

Chemistry
1 answer:
Scilla [17]3 years ago
7 0

The narrower the tube, the water will rise to a greater height.

Explanation:

Capillary rise is the height to which the water rises within the tube, and decreases as the width of the tube increases.

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Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh
Dima020 [189]

<u>Answer:</u> The \Delta G^o for the given reaction is -7.84\times 10^4J

<u>Explanation:</u>

For the given chemical reaction:

2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V  ( × 2)

<u>Reduction half reaction:</u> Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.36-(0.954)=0.406V

To calculate standard Gibbs free energy, we use the equation:

\Delta G^o=-nFE^o_{cell}

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

E^o_{cell} = standard cell potential = 0.406 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J

Hence, the \Delta G^o for the given reaction is -7.84\times 10^4J

4 0
3 years ago
What do cubic measurements, live cm3, mesure?"<br> O length<br> volume
anygoal [31]
The answer is volume.
6 0
3 years ago
The equilibrium constant, Kc, for the following reaction is 0.967 at 650 K. 2NH3(g) N2(g) 3H2(g) When a sufficiently large sampl
AlekseyPX

Answer: Concentration of NH_3 in the equilibrium mixture is 0.31 M

Explanation:

Equilibrium concentration of H_2 = 0.729 M

The given balanced equilibrium reaction is,

                 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

Initial conc.            x                0           0

At eqm. conc.     (x-2y) M     (y) M   (3y) M

The expression for equilibrium constant for this reaction will be:

3y = 0.729 M

y = 0.243 M

K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}

Now put all the given values in this expression, we get :

K_c=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}

0.967=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}

x=0.80

concentration of NH_3 in the equilibrium mixture = 0.80-2\times 0.243=0.31

Thus concentration of NH_3 in the equilibrium mixture is 0.31 M

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3 years ago
5 Examples of simple salts
kolbaska11 [484]
<h3>Five Examples of Salts </h3>

  • Sodium Chloride.
  • Sodium chloride (NaCl) is the most common type of salt in our lives.
  • Potassium Dichromate. •
  • Potassium dichromate (K2Cr2O7) is an orange-colored salt composed of potassium, chromium and oxygen.
  • Calcium Chloride.
  • Sodium Bisulfate.
  • Copper Sulfate.

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A few Critical Thinking questions/creativity on the topic 'Particles of matter are very small'
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Answer:

Any substance which occupies space and has mass is known as matter. States of matter: Matter exist in three different states that are Solid, Liquid and Gas. Matter is made up of very small particles

Explanation:

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