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2 years ago

Q24. List all of the ways you can force the reaction to shift to the product 5 points

Chemistry

1 answer:

densk [106]2 years ago

4 0

Answer:

1) Increase temperature

2) Decrease temperature

3) Increase concentration of reactants

4) Increase pressure

5) Decrease pressure

Explanation:

Le Chatelier's Principle Fundamentals states that a chemical reaction at equilibrium that undergoes changes to pressure, temperature, or concentration, this will cause the equilibrium to shift in the opposite direction to offset the change.

1) Increase temperature

2) Decrease temperature

3) Increase concentration of reactants

4) Increase pressure

5) Decrease pressure

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3 years ago

Binary Covalent (molecular) Compounds

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2 years ago

Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a

viva [34]

Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

$E^0_{[H^{+}/H_2]}=+0.00V$

$E^0_{[Ag^{+}/Ag]}=+0.799V$

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $H_2\rightarrow 2H^{+}+2e^-$

Reaction at cathode (reduction) : $Ag^{+}+1e^-\rightarrow Ag$

The balanced cell reaction will be,

$H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag$

Now we have to calculate the standard electrode potential of the cell.

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

$E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}$

$E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V$

Therefore, the standard cell potential will be +0.799 V

4 0

3 years ago

. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r

Tom [10]

Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

Explanation : Given,

Wavelength = $435.8nm=435.8\times 10^{-9}m$

conversion used : $1nm=10^{-9}m$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength = $435.8\times 10^{-9}m$

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}$

$E=4.56\times 10^{-19}J$

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

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3 years ago

A student weighed an empty graduated cylinder. It weighed 35.86 g. She then carefully added water to the graduated cylinder unti

Firdavs [7]

Question:

A student weighed an empty graduated cylinder. It weighed 35.86 g. She then carefully added water to the graduated cylinder until it reached the 7.5 mL mark. When she weighed the graduated cylinder again, this time with the 7.5 mL of water in it, it weighed 43.18 g. What was this student's experimental density of water?

Answer:

0.976 g/mL

Explanation:

Weight of empty cylinder = 35.86g

Volume of water = 7.5mL

Weight of cylinder + water = 43.18g

Experimental density = ?

Density of water = Mass of water / volume of water

Mass of water = (Weight of cylinder + water) - Weight of empty cylinder

Mass of water = 43.18 - 35.86 = 7.32g

Density = 7.32 / 7.5 = 0.976 g/mL

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2 years ago