What happens if more product is added to a system at equilibrium

Answers

Answer 1

Answer:

Answer:

More reactants will be produced

Explanation:

Le Chatelier's principle; adding additional product or reactant will move the equilibrium left or right to compensate and come back to equilibrium

By adding more product to your system at equilibrium, the equilibrium will shift towards reactants, more reactants will be produced

Answer 2

Answer:

Answer: Liquid molecules forming a gas and gas molecules forming a liquid are equal in number

Explanation: :/

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A chemist prepares hydrogen fluoride by means of the following reaction:CaF2 + H2SO4 --> CaSO4 + 2HF

The chemist uses 11 g of CaF2 and an excess of H2SO4, and the reaction produces 2.2 g of HF.
(a) Calculate the theoretical yield of HF.

(b) Calculate the percent yield of HF.

Answers

Answer:

39.3%

Explanation:

CaF2 + H2SO4 --> CaSO4 + 2HF

We must first determine the limiting reactant, the limiting reactant is the reactant that yields the least number of moles of products. The question explicitly says that H2SO4 is in excess so CaF2 is the limiting reactant hence:

For CaF2;

Number of moles reacted= mass/molar mass

Molar mass of CaF2= 78.07 g/mol

Number of moles reacted= 11g/78.07 g/mol = 0.14 moles of Calcium flouride

Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride

0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride

Mass of hydrogen fluoride formed (theoretical yield) = number of moles× molar mass

Molar mass of hydrogen fluoride= 20.01 g/mol

Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g ( theoretical yield of HF)

Actual yield of HF was given in the question as 2.2g

% yield of HF= actual yield/ theoretical yield ×100

%yield of HF= 2.2/5.6 ×100

% yield of HF= 39.3%

For the following reaction at equilibrium, which gives a change that will shift the position of equilibrium to favor formation of more products? 2NOBr(g) 2NO(g) + Br 2(g), ΔHº rxn = 30 kJ/mol

Answers

Answer:

Based on the given reaction, it is evident that the reaction is endothermic as indicated by a positive sign of enthalpy of reaction. Thus, it can be stated that the favoring of the forward reaction will take place by upsurging the temperature of the reaction mixture.

Apart from this, based on Le Chatelier’s principle, any modification in the quantity of any species is performed at equilibrium and the reaction will move in such an orientation so that the effect of the change gets minimized. Therefore, a slight enhancement in the concentration of the reactant will accelerate the reaction in the forward direction and hence more formation of the product takes place.

An element x is found to have a mass number of 31 and atomic number of 17. idenrify the group and the period to which it belongs?

Answers

Answer:

Well atomic number 17 is Chlorine, which is most commonly found as a gas, and is period 7.

Explanation:

elements found on period 7 are some of the most unstable elements.

Suppose of lead(II) acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the lead(II) acetate is dissolved in it. Round your answer to significant digits.

Answers

Answer:

0.294 M

Explanation:

The computation of the final molarity of acetate anion is shown below:-

Lead acetate = Pb(OAc)2

Lead acetate involves two acetate ion.

14.3 gm lead acetate = Mass ÷ Molar mass

= 14.3 g ÷ 325.29 g/mol

= 0.044 mole

Volume of solution = 300 ml.

then

Molarity of lead is

= 0.044 × 1,000 ÷ 300

= 0.147 M

Therefore the molarity of acetate anion is

= 2 × 0.147

= 0.294 M

Final answer:

To calculate the final molarity of acetate anion in the solution, consider the dissociation of lead(II) acetate and the presence of ammonium sulfate. When ammonium sulfate is added, it reacts with the lead(II) cations, leaving only the acetate anions in the solution. The final concentration of acetate anions is therefore the same as the initial concentration.

Explanation:

To calculate the final molarity of acetate anion in the solution, we need to consider the dissociation of lead(II) acetate and the presence of ammonium sulfate. Lead(II) acetate will dissociate into lead(II) cations (Pb2+) and acetate anions (CH3COO-) in solution. However, when ammonium sulfate is added, the sulfate anions (SO42-) react with the lead(II) cations, forming lead(II) sulfate and removing them from solution. This leaves us with only the acetate anions.

First, calculate the concentration of the acetate anions in the lead(II) acetate solution. Then subtract the concentration of the acetate anions that reacted with the lead(II) cations to form lead(II) sulfate. This will give us the final concentration of acetate anions in the solution.

Let's assume we have an initial concentration of lead(II) acetate of X M. The dissociation of lead(II) acetate can be represented as:

Pb(CH3COO)2(s) ⇌ Pb2+(aq) + 2CH3COO-(aq)

Since we assume the volume of the solution doesn't change when the lead(II) acetate is dissolved, the initial concentration of acetate anions is also X M.

When ammonium sulfate is added, it reacts with the lead(II) cations according to the reaction:

Pb2+(aq) + SO4^2-(aq) ⇌ PbSO4(s)

Since the concentration of lead(II) sulfate is negligible, we can assume that all the lead(II) cations react with the sulfate anions. This removes the lead(II) cations from solution, leaving us with only the acetate anions.

Therefore, the final concentration of acetate anions is still X M.

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How can you remove sand from salt? Which physical property of sand was used in the process?

Answers

Answer:

You could collect the mixture and pour it in water, stir it , ad filter out the sand. This uses the physical property of solubility.

Explanation:

The salt dissolved, the sand didn't.

At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.730 M

Answers

We have that from the Question, it can be said that The partial pressure of He would give a solubility of 0.730 M is

P_2=4.7atm

From the Question we are told

At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.730 M

Generally the equation for constant temperature is mathematically given as

$(C_2)/(C_1)=(P_2)/(P_1)\n\nTherefore\n\nP_2=(P_1C_1)/(C_1)\n\nP_2=(0.22*1.7)/(0.080)\n\nP_2=4.7atm\n\n$