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kherson [118]
2 years ago
15

Given the reaction: cu(s) + 4hno3(aq) → cu(no3)2(aq) + 2no3(g) + 2h2o(l )as the reaction occurs, what happens to copper?

Chemistry
1 answer:
katrin [286]2 years ago
6 0
From the reaction between Cu and HNO₃, the formed gas is NO₂ instead of NO₃. Hence the correct balanced equation would be,
Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O<span>(l)

Here, Cu goes to </span>Cu(NO₃)₂ by changing its oxidation number from 0 to +2 while NO₃⁻ goes to NO₂ by reducing its oxidation state from +5 to +4 . Hence Cu is oxidized by HNO₃ in the reaction.
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Which of the following atoms would have the longest de Broglie wavelength, if all have the same velocity?
GenaCL600 [577]

Answer:

Li

Explanation:

The phenomenon of wave particle duality was well established by Louis deBroglie. The wavelength associated with matter waves was related to its mass and velocity as shown below;

λ= h/mv

Where;

λ= wavelength of matter waves

m= mass of the particle

v= velocity of the particle

This implies that if the velocities of all particles are the same, the wavelength of matter waves will now depend on the mass of the particle. Hence; the wavelength of a matter wave associated with a particle is inversely proportional to the magnitude of the particle's linear momentum. The longest wavelength will then be obtained from the smallest mass of matter. Hence lithium which has the smallest mass will exhibit the longest DeBroglie wavelength

4 0
2 years ago
When a known quantity of compound, at a known concentration, is added to a known volume of another compound to determine the con
Vladimir [108]

Answer:

A titration

Explanation:

A common example of a titration is when we have an acid of unknown concentration, so we add a known volume of a base of known concentration. This process lets us determine the concentration of the acid.

By definition, a titration is a quantitative analysis, as we determine how much of an analyte is there in a sample. However, <u>there are quantitative analyzes which are not titrations</u>. This is why the most appropiate answer is<em> a titration</em>.

5 0
2 years ago
Alveolar air (a mixture of nitrogen, oxygen, and carbon dioxide) has a total pressure of 0.998 atm. If the partial pressure of o
inn [45]

Answer:

The partial pressure of carbon dioxide is 22.8 mmHg

Explanation:

Dalton's Law is a gas law that relates the partial pressures of the gases in a mixture. This law says that the pressure of a gas mixture is equal to the sum of the partial pressures of all the gases present.

In this case:

Ptotal=Pnitrogen + Poxygen + Pcarbondioxide

You know that:

  • Ptotal= 0.998 atm
  • Pnitrogen= 0.770 atm
  • Poxygen= 0.198 atm
  • Pcarbondioxide= ?

Replacing:

0.998 atm=0.770 atm + 0.198 atm + Pcarbondioxide

Solving:

Pcarbondioxide= 0.998 atm - 0.770 atm - 0.198 atm

Pcarbondioxide= 0.03 atm

Now you apply the following rule of three: if 1 atm equals 760 mmHg, 0.03 atm how many mmHg equals?

Pcarbondioxide=\frac{0.03 atm*760 mmHg}{1 atm}

Pcarbondioxide= 22.8 mmHg

<u><em>The partial pressure of carbon dioxide is 22.8 mmHg</em></u>

6 0
2 years ago
Watch the video to determine which of the following relationships correctly depict the relationship between pressure and volume
AnnZ [28]

Answer : The correct options are,

(B) V\propto \frac{1}{P}

(C) P\propto \frac{1}{V}

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

V\propto \frac{1}{P}

The relation between the pressure and volume of two gases are:

P_1V_1=P_2V_2

where,

P_1 = initial pressure of gas

P_2 = final pressure of gas

V_1 = initial volume of gas

V_2 = final volume of gas

5 0
2 years ago
A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (
djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

Kb = \frac{[ HSO3-][0H-]}{SO3-2}

Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation Kb=\frac{[HSO3-][OH-]}{[SO3-2]}

We noe have:

1.485*10^-^7=\frac{x*x}{(0.056-x)}

x = [OH-] = 9.11*10^-^5

pOH = -log(OH) = -log(9.11*10^-^5)

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0
2 years ago
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