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kobusy [5.1K]
1 year ago
6

What would the potential of a standard hydrogen electrode (s. h. e. ) be under the given conditions? [h ]=0. 84 mh2=2. 2 atm=298

k
Chemistry
1 answer:
Aneli [31]1 year ago
3 0

The potential of a standard hydrogen electrode (s. h. e. ) be under the given conditions is - 0.029 V

Calculation ,

Formula used :

E_{cell} = E_{cell} ° - 0.059/n ㏒H_{2}/ [H^{+}]^{2}

As we know that ,

[H^{+}] = 0. 84 M

P_{H_{2} = 2. 2 atm

Temperature ( T ) = 298  K

From reaction at electrode number of electrons involve ( n ) = 1

Standard electrode potential of  standard hydrogen electrode E_{cell} ° = 0

Putting the value of all data in equation ( i ) , we get

E_{cell} = 0  - 0.059/1 ㏒2. 2 / (0. 84)^{2} =  - 0.059 ㏒3.14 =  - 0.059×0.497

E_{cell} = - 0.029 V

To learn about electrode

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Mass number (A) of tungsten is 184

Now:

Atomic number of an element = Number of protons = Number of electrons

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In the case of tungsten

Total Number of electrons = 74

Jeffrey has already added 21 electrons

Number of electrons needed = 74 - 21 = 53

Ans (D) 53 more electrons are needed to complete the neutral atom of tungsten.

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3 years ago
The lighted half of the moon faces away from Earth during the _____ moon phase.
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The lighted half of the moon faces away from the earth during the New Moon phase
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3 years ago
Osmium is the most dense element we know of. A 22 g sample of Osmium has a volume of 100 cL. Calculate the density of Osmium in
kakasveta [241]

Answer:

a. V = 1000 mL

b. Denisty = 0.022 g/mL

Explanation:

a.

First we need to convert the volume of the Osmium into mL. For that purpose we are given the conversion unit as:

1 mL = 0.1 cL

Hence, the given volume of Osmium will be:

V = Volume of Osmium = 100 cL = (100 cL)(1 mL/0.1 cL) = 1000 mL

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b.

The density of Osmium is given by the following formula:

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5 0
3 years ago
A 1.800-g sample of solid phenol (C6H5OH(s)) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/?C. The temp
vichka [17]

Answer:

The balanced chemical equation:

C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)

Heat of combustion per gram of phenol is 32.454 kJ/g

Heat of combustion per gram of phenol is 3,050 kJ/mol

Explanation:

C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)

Heat capacity of calorimeter = C = 11.66 kJ/°C

Initial temperature of the calorimeter = T_1= 21.36^oC

Final temperature of the calorimeter = T_2= 26.37^oC

Heat absorbed by calorimeter = Q

Q=C\times \Delta T

Heat released during reaction = Q'

Q' = -Q ( law of conservation of energy)

Energy released on combustion of 1.800 grams of phenol = Q' = -(58.4166 kJ)

Heat of combustion per gram of phenol:

\frac{Q'}{1.800 g}=\frac{-58.4166 kJ}{1.800 g}=32.454 kJ/g

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Heat of combustion per gram of phenol:

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3 0
3 years ago
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tensa zangetsu [6.8K]
The correct answer is the fourth option. The complete dissociation of a strong base is BOH + h20 -> B+ + OH- + H20 since this is the only base from the choices given. A base is a substance that accepts hydrogen ions. 
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