What force(s)

egoroff_w [7]

Answer:

ionization constant of an acid (Ka) is the equilibrium constant for the ionization of an acid.

Explanation:

The acid ionization represents the fraction of the original acid that has been ionized in solution. Therefore, the numerical value of Ka is a reflection of the strength of the acid.

4 0

2 years ago

What is radio isotopes?

Evgesh-ka [11]

<span>Radioactive isotope, also called radioisotope, radionuclide, or radioactive nuclide, any of several species of the same chemical element with different masses whose nuclei are unstable and dissipate excess energy by spontaneously emitting radiation in the form of alpha, beta, and gamma rays.</span>

7 0

3 years ago

Read 2 more answers

How many grams of oxygen are needed to produce 5.40 g of water according to this equation? 2 H2 (g) + O2 (g) ⟶ 2 H2O (l)

garik1379 [7]

Explanation:

https://socratic.org/questions/using-the-equation-2-h2-o2-2-h2o-determine-how-many-grams-of-oxygen-will-be-need

8 0

2 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

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Does anyone know what Hydrogen's oxidation state would be if it was acting as an anion non-metal?

Ilia_Sergeevich [38]

Answer:

yeah,The oxidation state of an atom does not represent the "real" charge on that atom, or any other actual atomic property.Hydrogen has OS = +1, but adopts −1 when bonded as a hydride to metals or metalloids. Oxygen in compounds has OS = −2. This set of postulates covers .

Explanation:

3 0

2 years ago