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Viktor [21]
2 years ago
6

What's mole in 34g of H2O​

Chemistry
1 answer:
enyata [817]2 years ago
3 0

Answer:

I hope you have the answer now

Explanation:

u an

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Bromine vs Chlorine | Br vs Cl

 

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Bromine

Bromine is denoted by the symbol Br. This is in the 4th period of the periodic table between chlorine and iodine halogens. Its electronic configuration is [Ar] 4s2 3d10 4p5. The atomic number of bromine is 35. Its atomic mass is 79.904. Bromine staChlorine is an element in the periodic table which is denoted by Cl.  It is a halogen (17th group) in the 3rd period of the periodic table. The atomic number of chlorine is 17; thus, it has seventeen protons and seventeen electrons. Its electron configuration is written as 1s2 2s2 2p6 3s2 3p5. Since the p sub level should have 6 electrons to obtain the Argon, noble gas electron configuration, chlorine has the ability to attract an electron. ys as a red-brown color liquid at room temperature.

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G An example of an element is ____.
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You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.8
Lubov Fominskaja [6]

Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is pK_a  =7.82

Explanation:

From the question we are told

    The volume of base is  V_B = 26.mL = 0.0260L

     The pH of solution is  pH =  7.82

      The concentration of the acid is C_A = 0.1M

From the pH we can see that the titration is between a strong base and a weak acid

 Let assume that the the volume of acid is  V_A = 18mL= 0.018L

Generally the concentration of base

                    C_B = \frac{C_AV_A}{C_B}

Substituting value  

                     C_B = \frac{0.1 * 0.01800}{0.0260}

                    C_B= 0.0692M

When 13mL of the base is added a buffer is formed

The chemical equation of the reaction is

           HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}

Now before the reaction the number of mole of base is  

            No \ of \ moles[N_B]  =  C_B * V_B

Substituting value  

                    N_B = 0.01300 * 0.0692

                         = 0.0009 \ moles    

                                 

Now before the reaction the number of mole of acid is  

            No \ of \ moles  =  C_B * V_B

Substituting value  

                    N_A = 0.01800 *0.1

                         = 0.001800 \ moles

Now after the reaction the number of moles of  base is  zero  i.e has been used up

    this mathematically represented as

                         N_B ' = N_B - N_B = 0

    The  number of moles of acid is  

             N_A ' = N_A  - N_B

                   = 0.0009\ moles

The pH of this reaction can be mathematically represented as

                 pH  = pK_a + log \frac{[base]}{[acid]}

Substituting values

                  7.82 = pK_a +log \frac{0.0009}{0.0009}

                  pK_a  =7.82        

                     

             

                                 

       

           

                     

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Remember pH=-log(H ions). So it would be pH=-log(10^-7).
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Hello
Sulfur is the answer
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