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SashulF [63]
3 years ago
7

The total pressure of a confined mixture of gases is the sum of the pressures of each of the gases in the mixture. This is known

as ____. A. Charles' Law B. Tom's Law C. Boyle's Law D. Dalton's Law
Chemistry
1 answer:
olganol [36]3 years ago
5 0

Answer:

D

Explanation:

It is also known as the Dalton’s law of partial pressure. Given a confinement that contains a mixture of gases which do not mix, the total pressure equals the sum of the individual pressures.

The term, which do not mix is necessary because, if the gases are the type that mix, the law will no longer hold as they would have given up their individual identities and hence their individual partial pressure cannot be use to access them anymore.

Hence, the law helps to sum the totality of the pressures of a number of gases which exists together in a confinement and they do not mix. Say we have 3 gases A, B and C. The total pressure is the sum of pressure A, pressure B and pressure C.

You might be interested in
The ammonia molecule in the diagram has the observed bond orientation because ...
aivan3 [116]

Answer:

  • Nitrogen has four pairs of electrons: 3 bonds and 1 lone pair in the valence shell;
  • Electrons repel one another based on the VSEPR theory;
  • Nitrogen has a total of 7 protons (its atomic number is 7) in its nucleus.

Explanation:

The shape and the bond orientation of molecules and ions are both explained by the valences shell electron pair repulsion theory (VSEPR).

Ammonia, NH_3, is a molecule which contains three N-H bonds, as well as one lone pair on nitrogen. According to the VSEPR theory, molecules try to acquire a shape which would minimize the repulsion exhibited by the electron clouds present, that is, between the bonding (shared in a bond) and non-bonding (lone pair) electrons.

In VSEPR, our main step is to calculate the steric number, this is the sum of the number of bonds (ignoring the multiplicity of any bond) and the lone pairs on a central atom. In ammonia, we have 3 bonds and 1 lone pair, totaling to a steric number of 4. A steric number of 4 without any lone pairs on a central atom and just bonds would yield a tetrahedral shape with bond angles of 109.5^o.

Now, in this case, since we have a lone pair instead of a bond, it is repelling stronger decreasing the bond angles to about 107^o.

The greater the number of lone pairs, the lower the angle becomes.

To summarize:

  • Nitrogen has four pairs of electrons: 3 bonds and 1 lone pair in the valence shell;
  • Electrons repel one another based on the VSEPR theory;
  • Nitrogen has a total of 7 protons (its atomic number is 7) in its nucleus.
3 0
3 years ago
“Kids and grown-ups love it so, the happy world of HARIBO!” Who does not like the dark brown liquorice and its lovely taste?! Ta
KiRa [710]

Answer:

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3 0
2 years ago
Draw the structures of the major 1,2 and 1,4-products formed by reaction of 1 mole of Br2 with 3-methyl-2,4-hexadiene. Assume th
Marizza181 [45]

1234567891011121314151617181920

8 0
3 years ago
The following data were collected for the rate of disappearance of NO in the reaction 2NO(g)+O2(g)→2NO2(g)::
Anit [1.1K]

Answer:

a) The rate law is: v = k[NO]² [O₂]

b) The units are: M⁻² s⁻¹

c) The average value of the constant is: 7.11 x 10³ M⁻² s⁻¹

d) The rate of disappearance of NO is 0.8 M/s

e) The rate of disappearance of O₂ is 0.4 M/s

Explanation:

The experimental rates obtained can be expressed as follows:

v1 = k ([NO]₁)ᵃ ([O₂]₁)ᵇ = 1.41 x 10⁻² M/s

v2 = k ([NO]₂)ᵃ ([O₂]₂)ᵇ = 5.64 x 10⁻² M/s

v3 = k ([NO]₃)ᵃ ([O₂]₃)ᵇ = 1.13 x 10⁻¹ M/s

where:

k = rate constant

[NO]₁ = concentration of NO in experiment 1

[NO]₂ = concentration of NO in experiment 2

[NO]₃ = concentration of NO in experiment 3

[O₂]₁ = concentration of O₂ in experiment 1

[O₂]₂ = concentration of O₂ in experiment 2

[O₂]₃ = concentration of O₂ in experiment 3

a and b = order of the reaction for each reactive respectively.

We can see these equivalences:

[NO]₂ = 2[NO]₁

[O₂]₂ = [O₂]₁

[NO]₃ = [NO]₂

[O₂]₃ = 2[O₂]₂

So, v2 can be written in terms of the concentrations used in experiment 1 replacing [NO]₂ for 2[NO]₁ and [O₂]₂ by [O₂]₁ :

v2 = k (2 [NO]₁)ᵃ ([O₂]₁)ᵇ

If we rationalize v2/v1, we will have:

v2/v1 = k *2ᵃ * ([NO]₁)ᵃ * ([O₂]₁)ᵇ / k * ([NO]₁)ᵃ * ([O₂]₁)ᵇ (the exponent "a" has been distributed)

v2/v1 = 2ᵃ

ln(v2/v1) = a ln2

ln(v2/v1) / ln 2 = a

a = 2

(Please review the logarithmic properties if neccesary)

In the same way, we can find b using the data from experiment 2 and 3 and writting v3 in terms of the concentrations used in experiment 2:

v3/v2 = k ([NO]₂)² * 2ᵇ * ([O₂]₁)ᵇ / k * ([NO]₂)² * ([O₂]₂)ᵇ

v3/v2 = 2ᵇ

ln(v3/v2) = b ln 2

ln(v3/v2) / ln 2 = b

b = 1

Then, the rate law for the reaction is:

<u>v = k[NO]² [O₂]</u>

Since the unit of v is M/s and the product of the concentrations will give a unit of M³, the units of k are:

M/s = k * M³

M/s * M⁻³ = k

<u>M⁻² s⁻¹ = k </u>

To obtain the value of k, we can solve this equation for every experiment:

k = v / [NO]² [O₂]

for experiment 1:

k = 1.41 x 10⁻² M/s / (0.0126 M)² * 0.0125 M = 7.11 x 10³ M⁻² s⁻¹

for experiment 2:

k = 7.11 x 10³ M⁻² s⁻¹

for experiment 3:

k = 7.12 x 10³ M⁻² s⁻¹

The average value of k is then:

(7.11 + 7.11 + 7.12) x 10³ M⁻² s⁻¹ / 3 = <u>7.11 x 10³ M⁻² s⁻¹ </u>

The rate of the reaction when [NO] = 0.0750 M and [O2] =0.0100 M is:

v = k [NO]² [O₂]

The rate of the reaction in terms of the disappearance of NO can be written this way:

v = 1/2(Δ [NO] / Δt) (it is divided by 2 because of the stoichiometric coefficient of NO)

where (Δ [NO] / Δt) is the rate of disappearance of NO.

Then, calculating v with the data provided by the problem:

v = 7.11 x 10³ M⁻² s⁻¹ * (0.0750M)² * 0.0100M = 0.4 M/s

Then, the rate of disappearance of NO will be:

2v = Δ [NO] / Δt = <u>0.8 M/s</u>

The rate of disappearance of O₂ has to be half the rate of disappearance of NO because two moles of NO react with one of O₂. Then Δ [O₂] / Δt = <u>0.4 M/s</u>

With calculations:

v = Δ [O₂] / Δt = 0.4 M/s (since the stoichiometric coefficient is 1, the rate of disappearance of O₂ equals the rate of the reaction).

3 0
3 years ago
When an atom of oxygen (O) forms an anion by gaining two electrons, what is the anion’s charge?
Rudiy27

Answer:

bdndbdjdbdjdbdjdbsidbdidbsjsbsisbsidbd

7 0
3 years ago
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