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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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A student obtains the following data: Mass of empty, dry graduated cylinder: 21.577 g Volume added of NaCl solution: 4.602 mL Ma
klasskru [66]

<u>Answer:</u> The density of NaCl solution is 3.930 g/mL

<u>Explanation:</u>

We are given:

Mass of cylinder, m_1 = 21.577 g

Mass of NaCl and cylinder combined, M = 39.664 g

Mass of NaCl, m_2 = (M-m_1)=(39.664-21.577)g=18.087g

To calculate density of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

We are given:

Mass of NaCl = 18.087 g

Volume of NaCl solution = 4.602 mL

Putting values in above equation, we get:

\text{Density of NaCl}=\frac{18.087g}{4.602mL}\\\\\text{Density of NaCl}=3.930g/mL

Hence, the density of NaCl solution is 3.930 g/mL

7 0
3 years ago
Explain what makes a covalently bonded molecule polar. Use the molecules of CH4 and CH3F as examples in your explanation.
RideAnS [48]

Answer:

See explanation

Explanation:

The magnitude of electronegativity difference between atoms in a bond determines whether that bond will be polar or not.

If the electronegativity difference between atoms in a bond is about 1.7, the bond is ionic. If the electronegativity difference is greater than 0.4 and less than 1.7, the bond will have a polar covalent character. Lastly, if the electronegativity difference between the bond is less than or equal to 0.4, the covalent bond is non polar.

The electronegativity difference between carbon and hydrogen is about 0.4 which corresponds to a nonpolar covalent bond hence the molecule is nonpolar.

The electronegativity difference between carbon and fluorine is about 1.5 indicating a highly polar bond. This gives CH3F an overall dipole moment thereby making the molecule polar.

8 0
3 years ago
Which statement is true about SO4 2− ?
Nonamiya [84]
It is a poly atomic anion because it has more than one atom and a negative charge
4 0
3 years ago
Read 2 more answers
How many grams are in 562 dg?<br> 56.2 g<br> 56,200 g<br> 5.620 g<br> 5.62 g
lana66690 [7]

Answer:

56.2

Explanation:

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7 0
3 years ago
The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. combustion of
Olin [163]

Answer:

            Empirical Formula  =  C₃H₆O₁

Solution:

Data Given:

                      Mass of Ethyl Butyrate  =  3.61 mg  =  0.00361 g

                      Mass of CO₂  =  8.22 mg  =  0.00822 g

                      Mass of H₂O  =  3.35 mg  =  0.00335 g

Step 1: Calculate %age of Elements as;

                      %C  =  (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

                      %C  =  (0.00822 ÷ 0.00361) × (12 ÷ 44) × 100

                      %C  =  (2.277) × (12 ÷ 44) × 100

                      %C  =  2.277 × 0.2727 × 100

                      %C  =  62.09 %


                      %H  =  (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.00335 ÷ 0.00361) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.9279) × (2.02 ÷ 18.02) × 100

                      %H  =  0.9279 × 0.1120 × 100

                     %H  =  10.39 %


                      %O  =  100% - (%C + %H)

                      %O  =  100% - (62.09% + 10.39%)

                      %O  =  100% - 72.48%

                      %O  =  27.52 %

Step 2: Calculate Moles of each Element;

                      Moles of C  =  %C ÷ At.Mass of C

                      Moles of C  = 62.09 ÷ 12.01

                      Moles of C  =  5.169 mol


                      Moles of H  =  %H ÷ At.Mass of H

                      Moles of H  = 10.39 ÷ 1.01

                      Moles of H  =  10.287 mol


                      Moles of O  =  %O ÷ At.Mass of O

                      Moles of O  = 27.52 ÷ 16.0

                     Moles of O  =  1.720 mol

Step 3: Find out mole ratio and simplify it;

                C                                        H                                     O

             5.169                                10.287                              1.720

       5.169/1.720                       10.287/1.720                     1.720/1.720

               3.00                                   5.98                                   1

                  3                                      ≈ 6                                     1

Result:

         Empirical Formula  =  C₃H₆O₁

8 0
3 years ago
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