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mr_godi [17]
3 years ago
11

Jill mixes 4 grams of vanilla extract into the 800 grams of cake batter's he has prepared. What is the concentration of vanilla

in her "solution" of cake batter?
Chemistry
1 answer:
Ivenika [448]3 years ago
8 0

<u>Answer:</u> The concentration of vanilla extract in the solution is 0.50 % (m/m)

<u>Explanation:</u>

We are given:

Mass of vanilla extract (solute) = 4 grams

Mass of cake's batter (solvent) = 800 grams

Mass of solution = Mass of solute + Mass of solvent = 4 + 800 = 804 grams

To calculate the mass percentage (concentration) of vanilla extract in the solution, we use the equation:

\text{Concentration of vanilla extract}=\frac{\text{Mass of vanilla extract}}{\text{Mass of solution}}\times 100

Putting values in above equation, we get:

\text{Concentration of vanilla extract}=\frac{4g}{804g}\times 100\\\\\text{Concentration of vanilla extract}=0.50\%

Hence, the concentration of vanilla extract in the solution is 0.50 % (m/m)

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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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Oxygen gas generated in the thermal decomposition of potassium chlorate is collected over water. The volume of gas collected is
Leya [2.2K]

The mass of oxygen collected from the thermal decomposition of potassium chlorate at a temperature of 297 K and 762 mmHg is 0.16 g

<h3>How to determine the mole of oxygen produced </h3>

We'll begin by obtaining the number of mole of oxygen gas produced from the reaction. This can be obtained by using the ideal gas equation as illustrated below:

  • Volume (V) = 0.128 L
  • Temperature (T) = 297 K
  • Pressure (P) = 762 – 22.4 = 739.6 mmHg
  • Gas constant (R) = 62.363 mmHg.L/Kmol
  • Number of mole (n) =?

PV = nRT

739.6 × 0.128 = n × 62.363 × 297

Divide both sides by 62.363 × 297

n = (739.6 × 0.128) / (62.363 × 297)

n = 0.0051 mole

Thus, the number of mole of oxygen gas produced is 0.0051 mole

<h3>How to determine the mass of oxygen collected</h3>

Haven obtain the number of mole of oxygen gas produced, we can determine the mass of the oxygen produced as follow:'

  • Mole = 0.0051 mole
  • Molar mass of oxygen gas = 32 g/mole
  • Mass of oxygen =?

Mole = mass / molar mass

0.0051 = mass of oxygen / 32

Cross multiply

Mass of oxygen = 0.0051 × 32

Mass of oxygen = 0.16 g

Thus, we can conclude that the mass of oxygen gas collected is 0.16 g

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1 year ago
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