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Rufina [12.5K]
3 years ago
13

1. A chemist prepares hydrogen fluoride by means of the following reaction:

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
8 0

Answer:

a) <em>Theoretical Yield of HF = 5.64 grams</em>

b) <em>Percentage Yield = 39%</em>

Explanation:

Reaction Given:

CaF2 + H2SO4 -> CaSO4 + 2HF

CaF2 = 11g

H2SO4 = Used in excess

HF = 2.2 g production = Actual Yield

So, Let's write down the molar masses:

Molar Mass of CaF2 = 78 g /mol

Molar Mass of HF = 20 g/mol

From the reaction, we can see the 1 mole of CaF2 gives the 2 moles of HF

i.e

a) Theoretical Yield of HF:

1 mole CaF2 = 2 moles HF

78 g CaF2 = 2 x 20 g of HF

78 g CaF2 = 40 g of HF

1 g CaF2 = 40g/78g of HF

And in the question it is given that chemist used 11 g of CaF2 so,

1 x 11 g of CaF2 = 11 x 40/78 g of HF

11 g of CaF2 = 440/78 g of HF

11 g of CaF2 = 5.64 g of HF

And this is the theoretical yield

<em>Theoretical Yield of HF = 5.64 grams</em>

b) Now, calculate the Percentage Yield of HF

<em>Percentage Yield = Actual Yield /Theoretical Yield x 100</em>

Percentage Yield = 2.2 g /5.64 g x 100

Percentage Yield = 39%

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Answer:

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Explanation:

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Select the most likely product for this reaction: upper a g upper n upper o subscript 3 (a q) plus upper n a upper o upper h (a
Mamont248 [21]

One of the most likely products for the reaction would be Ag_2O

<h3>Chemical reactions</h3>

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2 AgNO_3 + 2 NaOH --- > Ag_2O + 2 NaNO_3 + H_2O

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5 0
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earnstyle [38]

Answer:

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Explanation:

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We need to find the heat energy absorbed in calories.

We know that the relation between joules and calories is as follows :

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What is the standard electrode potential for a galvanic cell constructed in the appropriate way from these two half-cells?
____ [38]

E

θ

Cell

=

+

2.115

l

V

Cathode

Mg

2

+

/

Mg

Anode

Ni

2

+

/

Ni

Explanation:

Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]

Mg

2

+

(

a

q

)

+

2

l

e

−

→

Mg

(

s

)

−

E

θ

=

−

2.372

l

V

Ni

2

+

(

a

q

)

+

2

l

e

−

→

Ni

(

s

)

−

E

θ

=

−

0.257

l

V

The standard reduction potential

E

θ

resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (

298

l

K

,

1.00

l

kPa

) is defined as

0

l

V

for reference. [2]

A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.

Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher

E

θ

and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower

E

θ

will experience oxidation and act the anode.

E

θ

(

Ni

2

+

/

Ni

)

>

E

θ

(

Mg

2

+

/

Mg

)

Therefore in this galvanic cell, the

Ni

2

+

/

Ni

half-cell will experience reduction and act as the cathode and the

Mg

2

+

/

Mg

the anode.

The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:

E

θ

cell

=

E

θ

(

Cathode

)

−

E

θ

(

Anode

)

E

θ

cell

=

−

0.257

−

(

−

2.372

)

E

θ

cell

=

+

2.115

Indicating that connecting the two cells will generate a potential difference of

+

2.115

l

V

across the two cells.

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2 years ago
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