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denpristay [2]
3 years ago
14

Which situation would cause the following equilibrium reaction to decrease the formation of the products? 2SO2 (g) + O2 (g) Two

arrows stacked on top of each other. The top arrow points to the right. The bottom arrow points to the left. 2SO3 (g) + Energy
increase the temperature
decrease the volume
increase the pressure
decrease the temperature
Chemistry
2 answers:
vovangra [49]3 years ago
8 0

The given equilibrium reaction is,

2 SO_{2} (g) + O_{2}(g)  2 SO_{3} (g) + Energy

The given reaction is exothermic. So, heat energy will be a product. Therefore, decreasing the temperature (heat energy) would lead to the formation of more products as when the amount of energy which is a product is reduced, there is more room for the products to form.

Increasing the pressure would shift the equilibrium towards that side which has least number of moles of the gaseous substance. Hence, here increasing the pressure would lead to the formation of more products by shifting the equilibrium towards the right side.

Decreasing the volume would make the equilibrium shift towards the least number of moles of the gaseous substance. So, here in this equilibrium decreasing the volume would lead to the formation of more products.

MaRussiya [10]3 years ago
8 0

Answer:  increase the temperature

Explanation: Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

2SO_2+O_2\leftrightharpoons 2SO_3+energy

This is a type of Exothermic reaction because heat is released in the reaction.

Option A: Increase the temperature

If the temperature is increased, so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in temperature occurs. As, this is an exothermic reaction, forward reaction will increase the temperature. Hence, the equilibrium will shift in the left direction, products will decrease.

Option B: Decrease the volume  and Option C: increase in pressure

If the volume of the container is decreased, the pressure will increase according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. As the number of moles of gas molecules is greater at the reactant side. So, the equilibrium will shift in the right direction, products will increase.

Option D: Decrease the temperature

If the temperature is decreased, so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in temperature occurs. As, this is an exothermic reaction, forward reaction will increase the temperature. Hence, the equilibrium will shift in the right direction, products will increase.

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Rank the following elements by effective nuclear charge, Zeff, for a valence electron. F LI Be B N
Stels [109]

Answer:

Rank in increasing order of effective nuclear charge:

  • Li < Be < B < N < F

Explanation:

This explains the meaning of effective nuclear charge, Zeff, how to determine it, and the calculations for a valence electron of each of the five given elements: F, Li, Be, B, and N.

<u>1) Effective nuclear charge definitions</u>

  • While the total positive charge of the atom nucleus (Z) is equal to the number of protons, the electrons farther away from the nucleus experience an effective nuclear charge (Zeff) less than the total nuclear charge, due to the fact that electrons in between the nucleus and the outer electrons partially cancel the atraction from the nucleus.

  • Such effect on on a valence electron is estimated as the atomic number less the number of electrons closer to the nucleus than the electron whose effective nuclear charge is being determined: Zeff = Z - S.

<u><em>2) Z eff for a F valence electron:</em></u>

  • F's atomic number: Z = 9
  • Total number of electrons: 9 (same numer of protons)
  • Period: 17 (search in the periodic table or do the electron configuration)
  • Number of valence electrons:  7 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 9 - 7 = 2
  • Zeff = Z - S = 9 - 2 = 7

<u><em>3) Z eff for a Li valence eletron:</em></u>

  • Li's atomic number: Z = 3
  • Total number of electrons: 3 (same number of protons)
  • Period: 1 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 1 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 3 - 1 = 2
  • Z eff = Z - S = 3 - 2 = 1.

<em>4) Z eff for a Be valence eletron:</em>

  • Be's atomic number: Z = 4
  • Total number of electrons: 4 (same number of protons)
  • Period: 2 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 2 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 4 - 2 = 2
  • Z eff = Z - S = 4 - 2 = 2

<u><em>5) Z eff for a B valence eletron:</em></u>

  • B's atomic number: Z = 5
  • Total number of electrons: 5 (same number of protons)
  • Period: 13 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 3 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 5 - 3 = 2
  • Z eff = Z - S = 5 - 2 = 3

<u><em>6) Z eff for a N valence eletron:</em></u>

  • N's atomic number: Z = 7
  • Total number of electrons: 7 (same number of protons)
  • Period: 15 (search on the periodic table or do the electron configuration)
  • Number of valence electrons: 5 (equal to the last digit of the period's number)
  • Number of electrons closer to the nucleus than a valence electron: S = 7 - 5 = 2
  • Z eff = Z - S = 7 - 2 = 5

<u><em>7) Summary (order):</em></u>

  Atom          Zeff for a valence electron

  • F                   7
  • Li                   1
  • Be                 2
  • B                   3
  • N                   5

  • <u>Conclusion</u>: the order is Li < Be < B < N < F
6 0
3 years ago
For a principal quantum number n, how many atomic orbitals are possible?
slavikrds [6]

Answer:

The total number of orbitals for a given n value is n2.

Explanation:

For a hydrogen atom with n=1, the electron is in its ground state; if the electron is in the n=2 orbital, it is in an excited state.

5 0
3 years ago
A+common+iv+solution+is+0.9%+saline+(nacl+solution).+what+is+the+osmolarity+of+0.9%+saline+mosmoles/l?+the+molecular+weight+of+n
Vedmedyk [2.9K]

An osmolarity of saline solution is 308 mosmol/L.

m(NaCl) = 9 g; the mass of sodium chloride

V(solution) = 1 L; the volume of the saline solution

n(NaCl) = 9 g ÷ 58.44 g/mol

n(NaCl) = 0.155 mol; the amount of sodium chloride

number of ions = 2

Osmotic concentration (osmolarity) is a measure of how many osmoles of particles of solute it contains per liter.

The osmolarity = n(NaCl) ÷ V(solution)  × 2

The osmolarity = 0.154 mol ÷ 1 L × 2

The osmolarity = 0.154 mol/L × 1000 mmol/m × 2

The osmolarity of the saline solution = 308 mosm/L.

More about osmolarity: brainly.com/question/13258879

#SPJ4

8 0
1 year ago
PLEASE HELPPP. I'LL AWARD BRAINLIEST !!!!!
dsp73

Answer:

the properties of catalyst are

it remains unchanged after chemical reactions

it accelerate or deaccelerate the reaction without taking part in it

Explanation:

they are used to convert raw materials into useful one

catalyst are integral in making plastics

5 0
3 years ago
A coffee-cup calorimeter initially contains 125 g water at 24.28C. Potassium bromide (10.5 g), also at 24.28C, is added to the w
iren2701 [21]

Answer:

The solution is given below

Explanation:

Heat, q= mc∆T

q= 125g x 4.18 J/g∙°C x (21.18x- 24.28) °C

q=  -1619.75J

NEGATIVE SIGN INDICATES THAT HEAT IS ABSORBED.

Enthalpy Change, ∆H = 1619.75 7/ 10.5 g

                                     = 154.26 J/g

No. of moles of KBr = Mass of KBr/ Molecular Weight of KBr

                                =10.5g/119gmol-1

                                =0.088 mol

∆H= 1619.75 J/ 0.088 mol

      = 18.41 kJ/mol  

6 0
3 years ago
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