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2 years ago

Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 15 ∘C.

Chemistry

1 answer:

umka2103 [35]2 years ago

8 0

Answer:

P(oxygen gas) = 3.75atm

P(helium) = 19.2atm

Ptotal = 22.95atm

Explanation:

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What is the oxidation number of chromium in dichromate ion, cr2o72-?

castortr0y [4]

Oxidation number is charge of element in compound. Can be neutral, positive or negative.
Oxygen in dichromate has oxidation number -2, becauce there are seven oxygens, net oxidation number of oxygen is -14.
2·x(oxidation number of Cr) + 7· (-2) = -2 2x= +12
x= +6, oxidation number of one chromium is +6.

8 0

3 years ago

Draw a line-bond structure for CBrN. Explicitly draw all H atoms. You do not have to include lone pairs in your answer. In cases

DochEvi [55]

Answer:

Br - C ≡ N

Explanation:

To draw the Lewis line-bond structure we need to bear in mind the octet rule, which states that in order to gain stability each <em>atom tends to share electrons until it has 8 electrons in its valence shell</em>.

C has 4 e⁻ in its valence shell so it will form 4 covalent bonds.

Br has 7 e⁻ in its valence shell so it will form 1 covalent bond.

N has 5 e⁻ in its valence shell so it will form 3 covalent bonds.

The most stable structure that respects these premises is:

Br - C ≡ N

It does not have any H atom.

8 0

2 years ago

Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

3 years ago

The decomposition of NH4HS is endothermic: NH4HS(s)⇌NH3(g)+H2S(g) Part A Which change to an equilibrium mixture of this reaction

shepuryov [24]

Explanation:

According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.

As the given reaction is as follows.

$NH_{4}HS(s) \rightleftharpoons NH_{3}(g) + H_{2}S(g)$

(a) When increase the temperature of the reactants or system then equilibrium will shift in forward direction where there is less temperature. It is possible for an endothermic reaction.

Thus, formation of $H_{2}S$ will increase.

(b) When we decrease the volume (at constant temperature) of given reaction mixture then it implies that there will be increase in pressure of the system. So, equilibrium will shift in a direction where there will be decrease in composition of gaseous phase. That is, in the backward direction reaction will shift.

Hence, formation of $H_{2}S$ will decrease with decrease in volume.

When we increase the mount of $NH_{4}HS$ then equilibrium will shift in the direction of decrease in concentration that is, in the forward direction.

Thus, we can conclude that formation of $H_{2}S$ will increase then.

3 0

3 years ago

Which of these elements is essential to making up all of the organic molecules?

siniylev [52]

Carbon

Hope this helps (:

5 0

3 years ago

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