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Darina [25.2K]
3 years ago
7

EASY QUESTION if you can have any superpower what would it be

Chemistry
2 answers:
Feliz [49]3 years ago
7 0

Answer:

well Itwould be the power to wish for anything

jolli1 [7]3 years ago
4 0

Answer:

fly like a bird

Explanation:

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Which part of the wave does the arrow point to?
AlekseyPX
I think it’s gonna be D
7 0
3 years ago
Two chemist are trying to determine if their water sample is a mixture or pure substance. In their investigation they first filt
Effectus [21]

The correct answer is option B. Dirty water is a mixture of solid particles and liquid. It is both a mixture and pure substance.

The dirty water sample has both gravel and liquid water in it. After filtration the gravel is removed so the water sample looks clearer than before filtration. Liquid water is a pure substance because it is a compound that is made up of elements hydrogen and oxygen. Now the gravel is only physically combined with the liquid water, thus giving the water sample properties of a mixture. And like any mixture, gravel is physically separated through filtration from the liquid water.

Thus the water sample of the chemists is both a mixture and pure substance.




8 0
4 years ago
Which species acts exclusively as spectator ions in this reaction
adelina 88 [10]

Answer:

K^+ and NO3^-

Explanation:

In a balanced ionic equation, we usually see the species that react to yield the main product in the reaction.

Consider the reaction;

Pb(NO3)2(aq) +2 KI(aq) -------> PbI2(s) + 2KNO3(aq)

The main product in this reaction is PbI2. Hence the balanced ionic equation is;

Pb^2+(aq) + 2I^-(aq) ------> PbI2(s)

Notice that K^+ and NO3^- did not participate in this reaction. All ions that are part of the molecular equation but do not participate in the ionic reaction equation are called spectator ions. Hence K^+ and NO3^- are spectator ions in this reaction as can be seen clearly above.

4 0
3 years ago
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        

                     

             

                                 

       

           

                     

8 0
3 years ago
The molarity of an aqueous sodium phosphate solution is 0.650 M. What is the molality of sodium ions present in this solution? T
netineya [11]

Answer:

molality of sodium ions is 1.473 m

Explanation:

Molarity is moles of solute per litre of solution

Molality is moles of solute per kg of solvent.

The volume of solution = 1 L

The mass of solution = volume X density = 1000mL X 1.43 = 1430 grams

The mass of solute = moles X molar mass of sodium phosphate = 0.65X164

mass of solute = 106.6 grams

the mass of solvent = 1430 - 106.6 = 1323.4 grams = 1.3234 Kg

the molality = \frac{moles of solute}{mass of solvent in kg}=\frac{0.65}{1.323}= 0.491m

Thus molality of sodium phosphate is 0.491 m

Each sodium phosphate of molecule will give three sodium ions.

Thus molality of sodium ions = 3 X 0.491 = 1.473 m

7 0
3 years ago
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