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yKpoI14uk [10]
3 years ago
6

Suppose there are two known compounds containing generic X and Y. You have a 1.00g sample of each compound. One sample contains

.34g of X and the other contains .44g of X. Identify plausible sets of formulas for the two compounds
Chemistry
1 answer:
natulia [17]3 years ago
6 0
<span>Both samples have a mass of 1 g. Suppose the first sample has a generic mass, X, of 0.34 then it follows that the mass of Y = 1 - 0.34 = 0.66g. So we have a mass ratio of 0.34 : 0.66 where our ratio is of the form X : Y. If we divide by the smallest mass we end up with 1 : 1.94. But Since 1.94 ~ 97/50 then we multiply through by 50, when we do, we end up with a chemical form of the form X50Y97. On the other hand if the sample has a mass of 0.44 of Y then it also follows that the mass of X = 1 - 0.44 = 0.56. So we have a mass ratio of 0.56 : 0.44. If we divide by the smallest mass we end up with 1.27 : 1. Since 1.27 ~ 127/100 then if we multiply through by 100, we end up with a chemical form X127Y100</span>
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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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<u>Answer:</u>

The correct answer option is B) 2.0 M.

<u>Explanation:</u>

We are given the number of grams of NaOH (Sodium Chloride) which are dissolved in 750 milliliters of water and we are to find its molarity.

We know the formula of molarity:

<em>Molarity =  (mass * 1000) / (volume * molecular mass)  </em>

Volume = 750 ml = 750 cm^3

Molecular mass = 40  

Mass = 60 grams

Substituting these values in the above formula:

Molarity = \frac{(60*1000)}{750*40} = 2.0 M

8 0
3 years ago
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Igoryamba

Answer:

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Explanation:

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Answer:

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its a synthesis

Explanation:

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3 years ago
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