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Mrrafil [7]
2 years ago
7

Propionic acid (CH3CH-COOH) has a K, 1.3 x 10^-5 and phosphoric acid (H3PO2) has a Ka = 7.5 x 10^-3

Chemistry
1 answer:
tatuchka [14]2 years ago
5 0

Answer:

D. CH3CH2COO- for CH3CH2COOH; H2PO4 - for H3PO4.

Explanation:

Hello!

In this case, since the ionization of a weak acid produces a conjugate acid, molecule to which the leaving H⁺ is added to, and a conjugate base, everything that remains after the withdrawal of the H⁺, we can note down the ionization of each acid down below:

CH_3CH_2COOH\rightleftharpoons CH_3CH_2COO^-+H^+\\\\H_3PO_4\rightleftharpoons H^++H_2PO_4^-

Thus, we notice that the answer is D. CH3CH2COO- for CH3CH2COOH; H2PO4 - for H3PO4 because the phosphoric acid is most likely to ionize to dihydrogen phosphate ions rather than phosphate, or hydrogen phosphate ions it undergoes a stepwise ionization in which the first step is the predominant one.

Best regards!

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Suppose now that you wanted to determine the density of a small crystal to confirm that it is phosphorus. From the literature, y
denis23 [38]

Answer:

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

Explanation:

Given that:

the density of the mixture = 1.82 g/mL

From the density of the pure samples

The density of CHCl_3 = 1.492 g/mL

The density of CHBr_3 = 2.890 g/mL

The total volume of the liquid mixture = 20.0 mL

Suppose the volume of  CHCl_3 = P ml

and the volume of CHBr_3 = Q ml

the sum of their volumes should be equal to the total volume of the mixture

P \ ml + Q \ ml = 20 ml  ----- (1)

However, we know that Density = mass/volume

∴ mass = density × volume

The equation can now be expressed as:

\mathtt{(Density \ of  \ CHCl_3 \times Vol. \ of  \ CHCl_3 ) + (Density  \  of \  CHBr_3  \times \ volume \ of \ CHBr_3)} = \mathtt{ (Density  \ of \ mixture \times volume \ of \ the \ mixture)}

1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL  ---- (2)

From equation (1) ;

let Q = 20 - P

The replace the value of P into equation (2)

1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL

1.492 P g + 57.8g - 2.890 P g =  36.4g

1.492 P g - 2.890 P g = 36.4g - 57.8g

-1.398 P g = -21.4g

P = -21.4g/-1.398g

P = 15.31 mL

Q = 20 - P

Q = (20 - 15.31) mL

Q = 4.69 mL

∴

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

6 0
3 years ago
A mixture of c2h2 and ch4 has a total mass of 230.9 g. when this mixture reacts completely with excess oxygen, the co2 and h2o p
Salsk061 [2.6K]
Interesting problem. Thanks for posting.

C2H2 + (3/2)02 ====>  H2O + 2CO2
CH4 +  2O2 =====> 2H2O + CO2

The molar mass of C2H2 = 2*12 + 2*1 = 26
The molar mass of CH4 = 1*12 + 4*1 = 16

The number of moles of C2H2 = x
The number of moles of  CH4 = y
26x + 16y = 230.9 grams

For water we get (from the C2H2). Water has a molar mass of 2*1 + 16 = 18

x*18 See the balanced equation to see what it is the same number of moles as C2H2
From the methane we get
y*18
2*y* 18. Again see the balanced equation to see where that 2 came from.
18x + 36y is the total amount of water. 

Now for the CO2. CO2 has a molar mass of 12 + 2*16 = 44
From C2H2  we get 2*44*x = 88x grams of CO2
From CH4 we get 1*y*44 grams of CO2
88x + 44y for CO2

Now we total to get the grand total of water and CO2
18x + 44y + 88x + 44y = 972.7 grams total.
106x + 88y =  972.7

Two equations, two unknowns, we should be able to solve this problem
26x + 16y = 230.9
106x + 88y = 972.7

I'm not going to go through the math unless you request me to do so. 
x = 8.03 moles
y = 1.38 moles 

The initial amount of C2H2 was 8.03 * 26 = 208.78
The initial amount of CH4 was 16*1.38 = 22.08
The total (as a check is 230.86 which is pretty close to the given amount.
So Methane's mass in the initial givens was 22.08 grams.
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What are two properties of most nonmetals?(1) high ionization energy and poor electrical conductivity(2) high ionization energy
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4 0
3 years ago
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