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UkoKoshka [18]
3 years ago
9

Please help I suck at chemistry

Chemistry
2 answers:
gayaneshka [121]3 years ago
8 0
C6h5oh would be the acid
shtirl [24]3 years ago
4 0
C6H5OH donated a hydrogen to the water, so the c6h5oh would be the acid
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Why does a beta particle have an atomic number of -1
Fudgin [204]
I cant entirely tell for now but an article on rodioactivity should solve the problem
6 0
2 years ago
A 32 L samples of xenon gas at 10°C is expanded to 35 L. Calculate the final temperature.
morpeh [17]

Answer:

              Final Temperature = 36.54 ⁰C

Explanation:

Lets suppose the gas is acting ideally, then according to Charle's Law, "<em>The volume of a fixed mass of gas at constant pressure is directly proportional to the absolute temperature</em>". Mathematically for initial and final states the relation is as follow,

                                                V₁ / T₁  =  V₂ / T₂

Data Given;

                  V₁  =  32 L

                  T₁  =  10 °C = 283.15 K             ∴ K = °C + 273.15

                  V₂  =  35 L

                  T₂  =  ??

Solving equation for T₂,

                         T₂  =  V₂ × T₁  / V₁

Putting values,

                         T₂  =  (35 L × 283.15 K) ÷ 32 L

                         T₂  =  309.69 K     ∴ ( 36.54 °C )

Result:

           As the volume is increased from 32 L to 35 L, therefore, the temperature must have increased from 10 °C to 36.54 °C.

3 0
3 years ago
Will the water level of the sea increase when there is global warming
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I hope this helps.  Let me know if anything is unclear.
8 0
3 years ago
Will give brainliest
sergiy2304 [10]
It shows the atomic number
8 0
2 years ago
If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

4 0
3 years ago
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