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2 years ago

[H3O+] = 3.6 x 10-2 M

Chemistry

1 answer:

spayn [35]2 years ago

7 0

Answer:

Explanation:

Hello there!

In this case, since this concentration of hydronium ions is given, we infer we may calculate that of the hydroxide ions, pH and pOH because no question was specifically given. That is why we proceed as follows:

- pH: Here, we use this concentration to calculate the pH according to the negative logarithm:

$pH=-log(3.6x10^{-2})=1.44$

Which means that the solution is acidic as the pH is less than 7.

- pOH: since the pH added to the pOH is 14, we can calculate the latter as follows:

$pOH=14-1.44=12.56$

- Hydroxide concentration: Given the pOH, we apply the antilogarithm to calculate such concentration as shown below:

$[OH^-]=10^{-pOH}=10^{12.56}=2.75x10^{-13}M$

Keep in mind, this is an arbitrary solution because no question was provided.

Regards!

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A thermometer shows that the temperature of the air in a room has increased.Which type of energy has increased?

liberstina [14]

Kinetic energy has increased.

The <em>Kinetic Molecular Theory</em> of gases states that the average kinetic energy of the molecules of a gas is directly proportional to its Kelvin temperature.

Thus, if the temperature of the air in a room has increased, the average kinetic energy of its molecules has increased.

5 0

3 years ago

Consider the following reactions and their respective equilibrium constants: NO(g)+12Br2(g)⇌NOBr(g)Kp=5.3 2NO(g)⇌N2(g)+O2(g)Kp=2

maxonik [38]

Answer:

Equilibrium constant of the given reaction is $1.3\times 10^{-29}$

Explanation:

$NO+\frac{1}{2}Br_{2}\rightleftharpoons NOBr$ .... $(K_{p})_{1}=5.3$

$2NO\rightleftharpoons N_{2}+O_{2}$ .... $(K_{p})_{2}=2.1\times 10^{30}$

The given reaction can be written as summation of the following reaction-

$2NO+Br_{2}\rightleftharpoons 2NOBr$

$N_{2}+O_{2}\rightleftharpoons 2NO$

......................................................................................

$N_{2}+O_{2}+Br_{2}\rightleftharpoons 2NOBr$

Equilibrium constant of this reaction is given as-

$\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}$

$=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})$

$=\frac{(K_{p})_{1}^{2}}{(K_{p})_{2}}$

$=\frac{(5.3)^{2}}{2.1\times 10^{30}}=1.3\times 10^{-29}$

7 0

2 years ago

Read 2 more answers

Suppose an aluminum- nuclide transforms into a phosphorus- nuclide by absorbing an alpha particle and emitting a neutron. Comple

AfilCa [17]

Explanation:

An alpha particles is basically a helium nucleus and it contains 2 protons and 2 neutrons.

Symbol of an alpha particle is $^{4}_{2}\alpha$ . Whereas a neutron is represented by a symbol $^{1}_{0}n$ , that is, it has zero protons and only 1 neutron.

Therefore, reaction equation when an aluminum- nuclide transforms into a phosphorus- nuclide by absorbing an alpha particle and emitting a neutron is as follows.

$^{27}_{13}Al + ^{1}_{0}n \rightarrow ^{30}_{15}P + ^{1}_{0}n$