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vichka [17]
2 years ago
4

Fill in the blanks with vocabulary terms.All vocab terms should be in lower case.

Chemistry
1 answer:
zepelin [54]2 years ago
3 0

Answer :

<u>Oxidation </u>is the loss of electrons.

<u>Reduction </u>is the gain of electrons.

The compound that became reduced acts as the <u>oxidizing </u>agent.

The compound that became oxidized acts as the <u>reducing </u>agent.

The measure of a compounds likeliness to gain or lose an electron is its <u>electrochemical potential</u> (E value).

A common electron carrier we will use a lot in this class is <u>NAD⁺</u> when it is in the oxidized state and <u>NADH </u>when it is in the reduced state.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced.

Electrochemical potential : It is defined as the measurement of the potential difference between two half cells.

Electron carrier : The molecules that are capable of accepting 1 or 2 electrons from one molecule and donating to another molecule in the process of electron transport.

There are two important electron carriers:

Nicotinamide adenine dinucleotide (NAD⁺ in its oxidized form and NADH in its reduced form).

Flavin adenine dinucleotide (FAD)

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A 24.1-g mixture of nitrogen and carbon dioxide is found to occupy a volume of 15.1 L when measured at 870.2 mmHg and 31.2oC. Wh
zmey [24]

Answer:

Mole fraction of nitrogen =   0.52

Explanation:

Given data:

Temperature =  31.2 °C

Pressure = 870.2 mmHg

Volume = 15.1 L

Mass of mixture = 24.1 g

Mole fraction of nitrogen = ?

Solution:

Pressure conversion:

870.2 /760 = 1.12 atm

Temperature conversion:

31.2 + 273 = 304.2 K

Total number of moles:

PV = nRT

n = PV/RT

n =  1.12 atm × 15.1 L / 0.0821 L.atm. mol⁻¹.K⁻¹ × 304.2 K

n = 16.9 L.atm.  /25 L.atm. mol⁻¹

n = 0.676 mol

Number of moles of nitrogen are = x

Then the number of moles of CO₂ = 0.676 - x

Mass of nitrogen = x mol . 28 g/mol and for CO₂ Mass = 44 g/mol ( 0.676 - x)

24.1  = 28x + ( 29.7 -44x)

24.1 - 29.7  =  28x  - 44x

-5.6 = -16 x

x = 0.35

Mole fraction of nitrogen:

Mole fraction of nitrogen = moles of nitrogen / total number of moles

Mole fraction of nitrogen =   0.35  mol / 0.676 mol

Mole fraction of nitrogen =   0.52

3 0
2 years ago
Perform the following calculations and give your answer with the correct number of significant figures:
Arte-miy333 [17]
When multiplying numbers, the term with the least significant digits gives how many significant digits will be in the answer. 2.995 has the least with 4 "sig figs", so the answer will have 4 significant digits as well:

2.995/0.16685 = <span>17.9502547198
</span>                           ↑↑  ↑<span>↑
                        4 sig figs

So the answer is 17.95.</span>
3 0
2 years ago
When 4.41g of phosphoric acid (H3PO4) react with 9.25g of barium hydroxide, water and insoluble barium phosphate form. [T/I-7] a
AnnZ [28]

Answer:

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

Explanation:

Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.

H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

We will balance it using the trial and error method.

First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

Finally, we will get the balanced equation by multiplying H₂O by 6.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

3 0
2 years ago
A gaseous compound is 30.4% nitrogen and 69.6% oxygen. a 6.06-gram sample of gas occupies a volume of 1.00 liter and exerts a pr
aksik [14]

A gaseous compound is 30.4% nitrogen and 69.6% oxygen by mass. A 5.25-g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 1.26 atm at -4.0°C. Which of the following is its molecular formula?  

1) NO2  

2) N3O6  

3) N2O5  

4) N2O4  

5) NO

8 0
2 years ago
For some hypothetical metal, the equilibrium number of vacancies at 600°C is 1 × 1025 m-3. If the density and atomic weight of t
makvit [3.9K]

Answer:

\frac{N_{v}}{N}=1.92*10^{-4}

Explanation:

First of all we need to find the amount of atoms per volume (m³). We can do this using the density and the molar mass.

7.40 \frac{g}{cm^{3}}*\frac{1mol}{85.5 g}*\frac{6.023*10^{23}atoms}{1mol}*\frac{1000000 cm^{3}}{1m^{3}}=5.21*10^{28}\frac{atoms}{m^{3}}

Now, the fraction of vacancies is equal to the N(v)/N ratio.

  • N(v) is the number of vacancies 1*10^{25}m^{-3}
  • N is the number of atoms per volume calculated above.

Therefore:  

The fraction of vacancies at 600 °C will be:

\frac{N_{v}}{N}=\frac{1*10^{25}}{5.21*10^{28}}  

\frac{N_{v}}{N}=1.92*10^{-4}

I hope it helps you!

 

7 0
2 years ago
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