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Scrat [10]
2 years ago
13

What stress will shift the following equilibrium system to the right? 2SO2(g) + O2(g) ⇌ 2SO3(g); ΔH= –98.8 kJ/mol Decreasing con

centration of SO2 Decreasing temperature Increasing concentration of SO3 Increasing volume
Chemistry
1 answer:
denpristay [2]2 years ago
3 0

Answer:

Decreasing temperature.

Explanation:

  • Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
  • The choices here:

1) Decreasing concentration of SO₂,

The concentration of  SO₂ decreases, so the reaction will be shifted to the left direction to suppress the decrease in SO₂ concentration.

2) Decreasing temperature,

ΔH= – 98.8 kJ/mol, the negative sign means that the reaction is exothermic and release the heat.

If the temperature decreased, this means that the products decrease, so the reaction will shift to the right to suppress the decrease in temperature.

<em>It is the right choice.</em>

3) Increasing concentration of SO₃:

Increasing concentration of SO₃ will shift the reaction to the left to suppress the effect of increasing the concentration of SO₃.

4) Increasing volume:

Increasing the volume will decrease the pressure.

Decreasing the pressure will shift the reaction of gases to the side of higher no. of moles (left side).

  • <em>So, the right choice is: Decreasing temperature.</em>

<em></em>

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Papessa [141]

Answer:

The equation for the reaction of one sodium bicarbonate ( NaHCO3 ) molecule with one citric acid (C6H8O7) molecule is the following:

Sodium Bicarbonate + Citric Acid ⇒ Water + Carbon Dioxide + Sodium Citrate

NaHCO3 + C6H8O7 ⇒ 3 CO2 + 3 H2O + Na3C6H5O7

Explanation:

The reaction is in balance, that is, the whole H2CO3 is not finished, but a little bit of this acid is left in the solution. Therefore, when sodium bicarbonate is added to the solution with citric acid, sodium citrate salt (C6H5O7Na3) and carbonic acid (H2CO3) are formed, which is rapidly broken down into water (H2O) and carbonic oxide (CO2).

C6H8O7 + NaHCO3 ⇒ C6H5O7Na3 + 3 H2CO3

C6H5O7Na3 + 3 H2CO3 ⇔ C6H5O7Na3 + 3 H2O + 3 CO2

5 0
2 years ago
Which of the following reactions are possible if sodium is more reactive than aluminum, 3 iron, and lead?
Deffense [45]

Answer:

Reaction I: Sodium + Aluminum chloride →Sodium chloride + Aluminum

Explanation:

Sodium being more reactive means that it will take the place of aluminium in whats called a displacement reaction and form.

Sodium chloride + Aluminum

8 0
2 years ago
Read 2 more answers
How many moles of sodium bicarbonate is needed to neutralize 0.8 ml of sulphuric acid?
svet-max [94.6K]

Answer:

n NaHCO3 = 9.6 E-3 mol

Explanation:

balanced reaction:

  • 2 NaHCO3(s) + H2SO4(ac) ↔ Na2SO4(ac) + 2 CO2(g) + 2 H2O(l)
  • assuming a concentration of H2SO4 6M....normally worked in the lab

⇒ n H2SO4 = 8 E-4 L * 6 mol/L = 4.8 E-3 mol H2SO4

according to balanced reaction, we have that for every mol of H2SO4 there are two mol of NaHCO3 ( sodium bicarbonate)

⇒ mol NaHCO3 = 4.8 E-3 mol H2SO4 * ( 2 mol NaHCO3 / mol H2SO4 )

⇒ ,mol NaHCO3 = 9.6 E-3 mol

So 9.6 E-3 mol NaHCO3,  are the minimun moles necessary to neutralize the acid.

6 0
2 years ago
What volume would 0.435 moles of hydrogen gas, h2, occupy at stp?
faust18 [17]
The conversion factor for volume at STP is \frac {1mol}{22.4L} or \frac {22.4L}{1mol}. Since we want volume, we would use \frac {22.4L}{1mol}. We conclude with the following calculations:

0.435mol H_{2} * \frac {22.4LH_{2}}{1molH_{2}} = 9.744L

The answer is 9.744L H2
3 0
2 years ago
Read 2 more answers
Complete and balance the following redox reaction in acidic solution As2O3(s) + NO3- (aq) → H3AsO4(aq) + N2O3(aq)
scoray [572]

Answer:

As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)

Explanation:

Oxidation: As_{2}O_{3}(s)\rightarrow H_{3}AsO_{4}(aq)

  • Balance As: As_{2}O_{3}(s)\rightarrow 2H_{3}AsO_{4}(aq)
  • Balance H and O in acidic medium: As_{2}O_{3}(s)+5H_{2}O(l)\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)
  • Balnce charge: As_{2}O_{3}(s)+5H_{2}O(l)-4e^{-}\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)......(1)

Reduction: NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)

  • Balance N: 2NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)
  • Balance H and O in acidic medium: 2NO_{3}^{-}(aq)+6H^{+}(aq)\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)
  • Balance charge: 2NO_{3}^{-}(aq)+6H^{+}(aq)+4e^{-}\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)......(2)

Equation (1)+Equation (2) gives-

As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)

3 0
2 years ago
Read 2 more answers
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