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Afina-wow [57]
3 years ago
6

What elements are in KHCO3

Chemistry
1 answer:
Nutka1998 [239]3 years ago
5 0
Hydrogen H 1.00794
Carbon C 12.0107
Oxygen O 15.9994
Potassium K 39.0983
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How does a purebred individual differ from a hybrid individual?
zimovet [89]
Im gonna go based on the content of their blood. So as the generation continues, a purebred's blood has not differentiated too much and contains 98-100% of a "pure" bloodline meaning no outsiders of where this purebred originated has mixed. A hybrid is where the original bloodline is mixed with many others, so while a purebred has 99% of the original bloodline and the hybrid has 50% and he other half belongs to an entirely different line.
5 0
3 years ago
Calculate the masses of oxygen and nitrogen that are dissolved in of aqueous solution in equilibrium with air at 25 °C and 760 T
shtirl [24]

Explanation:

Let us assume that the volume of given aqueous solution is 7.5 L.

Therefore, according to Henry's law, the relation between concentration and pressure is as follows.

                C = \frac{P}{K_{h}}

where,   pressure (P) = 760 torr = 1 atm

According to Henry's law, constants for gases in water at 25^{o}C are as follows.

  p(O_{2}) = 0.21 atm = 0.21 bar

   p(N_{2}) = 0.78 atm = 0.78 bar

   K_{h} for O_{2} = 7.9 \times 10^{2} bar/mol

    K_{h} for N_{2} = 1.6 \times 10^{3} bar /mol

Since, 21% oxygen is present in air so, its mass will be 0.21 g. Similarly, 78% nitrogen means the mass of nitrogen is 0.78 g.

Therefore, concebtrations will be calculated as follows.

      C(O_{2}) = \frac{0.21}{7.9 \times 10^{2}} = 2.66 \times 10^-4 mol/L  

      C(N_2) = \frac{0.78}{1.6 \times 10^3} = 4.875 \times 10^-4 mol/L

Now, we will calculate the number of moles as follows.

         n(O_{2}) = 7.5 \times 2.66 \times 10^-4 = 1.995 \times 10^-3 mol

         n(N_2) = 7.5 \times 4.875 \times 10^-4 = 3.66 \times 10^-3 mol

As the molar mass of O_2 = 32 g/mol

Hence, mass of oxygen will be as follows.

         Mass of O_2 = 32 \times 1.995 \times 10^-3 \times 1000

                           = 63.84 mg

As the molar mass of N_{2} = 28

       Mass of N_{2} = 28 \times 3.66 \times 10^-3 = 102.5 mg

Thus, we can conclude that mass of oxygen is 63.84 mg  and nitrogen is 102.5 mg.

5 0
3 years ago
Read 2 more answers
Balance the equation for ethane C3H6 burning in oxygen to form carbon dioxide and steam. _____C3H6 + ____ O2 ---> CO2 + ____H
aleksandrvk [35]

Answer:

   2C3H6 + 9 O2 ---> 6 CO2 + 6 H2O

Explanation:

8 0
3 years ago
Which separation techniques will you apply for the separation of the following? a) oil from water b) kerosene and petrol c) salt
REY [17]

Answer:

A) Separating funnel method

B) Simple Distillation

C) Evaporation

D) Sublimation

E) It is based on the principle of separation whereby even though two substances are dissolved in the same solvent, their respective solubilities could be different. Thus, the component that has more solubility will rise fastest and will therefore get separated from the mixture.

Explanation:

A)

B) Kerosene and petrol are both miscible liquids and the difference in their boiling point temperature is not more than 25°C. Thus, we make use of Simple distillation.

C) Can be separated by evaporation where the water is boiled and it evaporates and leaves the salt behind

D) To separate camphor from salt, we use sublimation so the camphor can change directly from solid to the gas state without passing through the liquid state.

E) Chromatography is used to separate components of a mixture.

It is based on the principle of separation whereby even though two substances are dissolved in the same solvent, their respective solubilities could be different. Thus, the component that has more solubility will rise fastest and will therefore get separated from the mixture.

7 0
3 years ago
What is the ratio of [a–]/[ha] at ph 2.75? the pka of formic acid (methanoic acid, h–cooh) is 3.75?
Varvara68 [4.7K]

The dissociation of formic acid is:

HCOOH \rightleftharpoons HCOO^{-} + H^{+}

The acid dissociation constant of formic acid, k_a is:

k_a = \frac{[HCOO^{-}]  [H^{+}]}{HCOOH}

Rearranging the equation:

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}

pH = 2.75

pH = -log[H^{+}]

[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}

pk_a = 3.75

k_a = 10^{-3.75} = 1.78\times 10^{-4}

Substituting the values in the equation:

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}

Hence, the ratio is \frac{1}{10}.

4 0
3 years ago
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