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ddd [48]
3 years ago
7

What, if any, experimental evidence do you have that the equilibrium is affected by the addition of NaOH? How is this in accorda

nce with Le Châtelier’s principle?
Chemistry
1 answer:
vitfil [10]3 years ago
7 0
Sodium Hydroxide (NaOH) is also known as lye which is a base (very high ph; Alkaline)

Now, in chemistry, equilibrium is what affects the reaction rate of a reaction.  If they are in equilibrium, the concentrations of them will not change (both reactants and products).

Now, lets say that to synthesize a certain chemical, we need it to be in an acidic environment with HCL or some other acid as the catalyst for the reaction.

Well, if we were to add Sodium Hydroxide to this which is very alkaline, the ph would change greatly which affects the reaction rate.  If we do not have enough energy to overcome the activation barrier, the reaction will not occur (atleast for a very long time).

However, a common mistake is thinking that a catalyst will affect the equilibrium.  This is not true.  The reaction will still take place but it will have a very slow reaction rate.

TLDR; Adding a catalyst (like NaOH or Sodium Hydroxide) will not change the equilibrium but instead change the reaction rate.  The reaction can still occur, although it can take a very, very long time (like diamonds turning into graphite)


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Which of the following has potential energy but no kinetic energy? Longitudinal sound waves An arrow shot from a bow A compresse
Vikentia [17]

Answer:

A compressed spring

Explanation:

A compressed spring has potential energy only and no kinetic energy.

This is because kinetic energy is only possessed by particles in motion.

Energy in a compressed spring= -1/2kx² where x is the displacement.

In this equation there is no velocity so there is no kinetic energy.

3 0
3 years ago
Read 2 more answers
The enthalpy of a pure liquid at 75oC is 100 J/mol. The enthalpy of the pure vapor of that substance at 75oC is 1000 J/mol. What
pentagon [3]

Answer:

900 J/mol

Explanation:

Data provided:

Enthalpy of the pure liquid at 75° C = 100 J/mol

Enthalpy of the pure vapor at 75° C = 1000 J/mol

Now,

the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.

Thus, mathematically,

The heat of vaporization at 75° C

=  Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C

on substituting the values, we get

The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol

or

The heat of vaporization at 75° C = 900 J/mol

8 0
3 years ago
Consider the formation of hcn by the reaction of nacn (sodium cyanide) with an acid such as h2so4 (sulfuric acid): 2nacn(s)+h2so
Artemon [7]

2NaCN(s) + H₂SO₄(aq) --> Na₂SO₄(aq) + 2HCN(g)  

The molar ratio between NaCN : HCN is 2:2  or 1:1

Mass of HCN = 16.7 g

Molar mass of HCN = 1 + 12 + 14 = 27 g/mol

Molar mass of NaCN = 49 g/mol

Therefore, the mass of NaCN is

16.7 g of  HCN x 49 g/mol of NaCN / 27 g/mol of HCN = 30.3 grams of NaCN

Therefore, 30.3 grams of NaCN gives the lethal dose in the room.

5 0
3 years ago
Read 2 more answers
Please help!!
hichkok12 [17]
It’s 65% average that your answer
6 0
2 years ago
Suppose a 10.0 mL sample of an unknown
mote1985 [20]

The concentration of HCl is equal to 2.54mol/L.

<h3>Mole calculation</h3>

To solve this question, one must use the molarity calculation, which corresponds to the following expression:

                                               M = \frac{mol}{v}

Thus, to find the molarity of the sample, the following calculations must be performed:

V_f = 10ml + 625ml = > 635ml

                                              \frac{0.004mol}{xmol} =\frac{1000ml}{635ml}

                                                 x = 0.00254 mol

So, 0.00254 moles were added per 10ml, so we can do:

                                              \frac{0.00254mol}{xmol}= \frac{10ml}{1000ml}  \\x = 2.54mol/L

So, the concentration of HCl is equal to 2.54mol/L.

Learn more about mole calculation in: brainly.com/question/2845237

6 0
2 years ago
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