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anastassius [24]
3 years ago
12

Consider the following system at equilibrium: P(aq)+Q(aq)⇌3R(aq) Classify each of the following actions by whether it causes a l

eftward shift, a rightward shift, or no shift in the direction of the net reaction. Drag the appropriate items to their respective bins.
Items:1) Increase [P]2) Increase [Q]3) Increase [R]4) Decrease [P]5) Decrease [Q]6) Decrease [R]7) Triple [P] and reduce [Q] to one third8) Triple both [Q] and [R]
Chemistry
1 answer:
Dahasolnce [82]3 years ago
3 0

Explanation:

P(aq)+Q(aq)⇌3R(aq)

This problem involves applying LeChatelier's principle.

LeChatelier's principle states that whenever a system in equilibrium is disturbed, the equilibrium position would change in order to annul that change.

1) Increase [P]

This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.

2) Increase [Q]

This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.

3) Increase [R]

This would cause the equlibrium position to shift to the left. This is because more products have been formed, to annul that change more reactants have to be formed.

4) Decrease [P]

This would cause the equlibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.

5) Decrease [Q]

This would cause the equilibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.

6) Decrease [R]

This would cause the equilibrium position to shift to the right. This is because there are now less products, to annul that change more products have to be formed.

7) Triple [P] and reduce [Q] to one third

No shift in the direction of the net reaction because both changes cancels each other.

8) Triple both [Q] and [R]

No shift in the direction of the net reaction because both changes cancels each other.

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4NH3 + 502 --> 4NO + 6H20
gregori [183]

Answer:

5.52 g

Explanation:

  • 4NH₃ + 5O₂ → 4NO + 6H₂O

First we <u>convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:

  • 6.30 g NH₃ ÷ 17 g/mol = 0.370 mol NH₃
  • 1.80 g O₂ ÷ 32 g/mol = 0.056 mol O₂

Now we <u>calculate with how many NH₃ moles would 0.056 O₂ moles react</u>, using the<em> stoichiometric coefficients</em>:

  • 0.056 mol O₂ * \frac{4molNH_3}{5molO_2} = 0.045 mol NH₃

As there more NH₃ moles than required, NH₃ is the excess reactant.

Then we calculate how many NH₃ moles remained without reacting:

  • 0.370 mol NH₃ - 0.045 mol NH₃ = 0.325 mol NH₃

Finally we convert NH₃ moles into grams:

  • 0.325 mol NH₃ * 17 g/mol = 5.52 g
7 0
2 years ago
NEED HELP FAST I WILL MARK BRAINLIEST AND THANK YOU AND RATE 5 STARS.
sleet_krkn [62]
Hello there,

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Given 50g of sodium how many moles do you have? ​
Anastaziya [24]

Answer:

Amount of Na = 2.17moles

Explanation:

Mass of Na = 50g

Molar mass of Na = 23g/mol

Amount of mole = mass/molar mass

Amount of mole = 50/23

Amount of mole = 2.17moles

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3 years ago
g If 50.0 mL of a 0.75 M acetic acid solution is titrated with 1.0 M sodium hydroxide, what is the pH after 10.0 mL of NaOH have
V125BC [204]

Answer:

pH = 2.66

Explanation:

  • Acetic Acid + NaOH → Sodium Acetate + H₂O

First we <u>calculate the number of moles of each reactant</u>, using the <em>given volumes and concentrations</em>:

  • 0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid
  • 1.0 M NaOH * 10.0 mL = 10 mmol NaOH

We<u> calculate how many acetic acid moles remain after the reaction</u>:

  • 37.5 mmol - 10 mmol = 27.5 mmol acetic acid

We now <u>calculate the molar concentration of acetic acid after the reaction</u>:

27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M

Then we <u>calculate [H⁺]</u>, using the<em> following formula for weak acid solutions</em>:

  • [H⁺] = \sqrt{C*Ka}=\sqrt{0.458M*1.76x10^{-5}}
  • [H⁺] = 0.0028

Finally we <u>calculate the pH</u>:

  • pH = -log[H⁺]
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