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kirill [66]
3 years ago
7

Consider the reaction that occurs when copper is added to nitric acid. Upper C u (s) plus 4 upper H upper N upper O subscript 3

(a q) right arrow upper C u (upper N upper O subscript 3) subscript 2 (a q) plus 2 upper N upper O 2 (g) plus 2 upper H 2 upper o (l). What is the reducing agent in this reaction?
Chemistry
1 answer:
Nitella [24]3 years ago
8 0

Answer:

Cu

Explanation:

In the given reaction of the addition of copper to nitric acid,

Cu(s) + 4HNO3(aq) -> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

Cu or copper would be characterized as the reducing agent in this reaction. It is the chemical substance that is losing electrons and being oxidized due to this reduction/loss in this redox reaction as it is the metal that loses electrons by reacting with the non-metals.

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What types of fields does moving electricity produce?
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3 0
3 years ago
Alchemists in the Middle Ages dreamed of converting base metals, such as lead, into precious metals—gold and silver. Why could t
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Explanation:

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Transmutation of one chemical element into the another involves the changing of the atomic number. Such nuclear reaction requires millions of the times more energy as compared to normal chemical reactions. Thus, the dream of  the alchemist of transmuting the lead into the gold was never achievable chemically .

Conversion of lead to gold in today's world:

This conversion is indeed possible. The requirements are a particle accelerator, tremendous supply of the energy. Nuclear scientists at the Lawrence Berkeley National Laboratory located in California, more than 30 years ago, succeeded in producing very minute amounts of the gold from the bismuth. Bismuth is a metallic element which is adjacent to the lead on periodic table. Same process would work for the lead but isolating gold at end of reaction would prove much more difficult because lead is available in many isotopes. The homogeneous nature of the element means that it is easier to separate the gold from the bismuth as compared to separate the gold from the lead which has four  isotopic identities which all are stable.

4 0
4 years ago
After a polypeptide chain has been synthesized, certain amino acids in the peptide may become modified. For each modified amino
AnnZ [28]

Answer:

The modification that yields the first amino acid depicted is lysine

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5 0
3 years ago
Read 2 more answers
The equilibrium constant for the reaction
Hitman42 [59]

The question is incomplete, here is the complete question:

The equilibrium constant for the reaction

N₂O₄(g)⇌2NO₂ at 2°C is Kc = 2.0

If each yellow sphere represents 1 mol of N₂O₄(g) and each gray sphere 1 mol of NO₂ which of the following 1.0 L containers represents the equilibrium mixture at 2°C?

The image is attached below.

<u>Answer:</u> The system which represents the equilibrium having value of K_c=2.0 is system (b)

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c

For a general chemical reaction:

aA+bB\rightarrow cC+dD

The expression for K_{c} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

For the given chemical equation:

N_2O_4(g)\rightleftharpoons 2NO_2

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}      .......(1)

We are given:

Volume of the container = 1.0 L

Value of K_c = 2.0

Molarity of the substance is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}

For the given images:

  • <u>For a:</u>

Number of Gray spheres = 8 moles

Number of yellow spheres = 4 moles

Putting values in expression 1, we get:

K_c=\frac{(8/1)^2}{(4/1)}\\\\K_c=16

  • <u>For b:</u>

Number of Gray spheres = 4 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

K_c=\frac{(4/1)^2}{(8/1)}\\\\K_c=2

  • <u>For c:</u>

Number of Gray spheres = 6 moles

Number of yellow spheres = 6 moles

Putting values in expression 1, we get:

K_c=\frac{(6/1)^2}{(6/1)}\\\\K_c=6

  • <u>For d:</u>

Number of Gray spheres = 2 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

K_c=\frac{(2/1)^2}{(8/1)}\\\\K_c=\frac{1}{2}

Hence, the system which represents the equilibrium having value of K_c=2.0 is system (b)

3 0
3 years ago
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