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mina [271]
1 year ago
12

in the reaction: mg(s) 2hcl(aq) → mgcl2(aq) h2(g) how does the equilibrium shift if the hcl concentration is increased? to the p

roducts no change will occur to the reactants the equilibrium will initially shift but eventually be uneffected.
Chemistry
1 answer:
dedylja [7]1 year ago
4 0

The equilibrium shift is if the HCl concentration is increased in the product.

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

<h3>What is the effect of concentration on equilibrium?</h3>

If the concentration of a substance is changed, then the equilibrium will shift in such a way that it minimizes the effect of change that occurs. If we increase the concentration of a reactant, then the equilibrium will shift to minimize the changes in the direction of the reaction which uses the reactants, so that the reactant concentration decreases.

<h3>Factors affecting the concentration of the reaction</h3>
  • The temperature: As we increase the temperature, the average speed of the reactant molecules also increases. As many molecules move faster, a large number of molecules moving fast enough to react increases, making the faster formation of products.
  • pressure
  • and concentration of the system is the factors that affect equilibrium.

Thus, we concluded that with an increase in the concentration of reactant equilibrium shifts forward.

Learn more about equilibrium here:-

brainly.com/question/12252562

#SPJ4

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Because it went through a chemical change which changes its atomic form

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica
FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = \frac{1}{2}\times 0.52=0.26mol of lead (II) iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0
3 years ago
Which of the following most likely happens when the volume of a gas increases?
mafiozo [28]

the pressure of the gas increases

6 0
3 years ago
What volume (in L) of oxygen will be required to produce 77.4 L of water vapor in the reaction below?
Inga [223]

Answer:

90.3 L

Explanation:

Given data:

Volume of water produced = 77.4 L

Volume of oxygen required = ?

Solution:

Chemical equation:

2C₂H₆ + 7O₂  →  4CO₂ + 6H₂O

It is known that,

1 mole = 22.414 L

There are 7 moles of oxygen = 7×22.414 = 156.9 L

There are 6 moles of water = 6×22.414 = 134.5 L

Now we will compare:

                               H₂O           :              O₂    

                               134.5         :              156.9

                                 77.4         :             156.9/134.5×77.4 =90.3 L

So for the production of 77.4 L water 90.3 L oxygen is required.

8 0
3 years ago
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