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3 years ago

9

At room temperature, the solubility of which solute in water would be most affected by a change in pressure?

2 answers:

Nezavi [6.7K]3 years ago

7 0

Answer:

Carbon dioxide

Explanation:

Carbon dioxide is a product derived from the reaction of different processes, such as: the combustion of coal and hydrocarbons, the fermentation of liquids and the breathing of humans and animals. It is also in low concentration in the Earth's atmosphere.

The carbon dioxide structure is made up of molecules of linear geometry and non-polar character. In addition, at room temperature, the carbon dioxide solubility would be affected if a pressure change occurred.

Bad White [126]3 years ago

4 0

<h3><u>Answer;</u></h3>

At room temperature, the solubility of which solute in water would be most affected by a change in pressure?

<h3><u>Explanation;</u></h3>

As the pressure on a gas confined above a liquid increases, the solubility of the gas increases in the liquid.
Carbon dioxide gas is most soluble in water under conditions of high pressure and low temperature.
The solubility of gases decreases with increasing temperature, so the air space inside the cylinder when cold water is used will be less compared to warm water.

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The element in the top left corner of the periodic table is ______________.answer:

rodikova [14]

The answer is hydrogen

5 0

2 years ago

Nitrogen dioxide gas is dark brown in color and remains in equilibrium with dinitrogen tetroxide gas, which is colorless.

sukhopar [10]

Answer:

This reaction is exothermic because the system shifted to the left on heating.

Explanation:

2NO₂ (g) ⇌ N₂O₄(g)

Reactant => NO₂ (dark brown in color)

Product => N₂O₄ (colorless)

From the question given above, we were told that when the reaction at equilibrium was moved from room temperature to a higher temperature, the mixture turned dark brown in color.

This simply means that the reaction does not like heat. Hence the reaction is exothermic reaction.

Also, we can see that when the temperature was increased, the reaction turned dark brown in color indicating that the increase in the temperature favors the backward reaction (i.e the equilibrium shift to the left) as NO₂ which is the reactant is dark brown in color. This again indicates that the reaction is exothermic because an increase in the temperature of an exothermic reaction will shift the equilibrium position to the left.

Therefore, we can conclude that:

The reaction is exothermic because the system shifted to the left on heating.

8 0

2 years ago

The measured voltage of an electrochemical cell consisting of pure nickel immersed in a solution of Ni2+ ions of unknown concent

Verizon [17]

Answer:

[Ni²⁺] = 1.33 M

Explanation:

To do this, we need to use the Nernst equation which (in standard conditions of temperature)

E = E° - RT/nF lnQ

However R and F are constant, and the reaction is taking place in 25 °C so we can assume the nernst equation like this:

E = E° - 0.05916/n logQ

As the nickel is in the cathode, this means that this element is being reducted while Cadmium is being oxidized, therefore the REDOX reaction would be:

Cd(s) + Ni²⁺(aq) --------> Cd²⁺(aq) + Ni(s)

With this, Q:

Q = [Cd²⁺] / [Ni²⁺]

Now, we need to know the value of the standard reduction potentials, which can be calculated with the semi equations of reduction and oxidation:

Cd(s) ------------> Cd²⁺ + 2e⁻ E°₁ = 0.40 V

Ni²⁺ + 2e⁻ -------------> Ni(s) E°₂ = -0.25 V

E° = E°₁ + E°₂

E° = 0.40 - 0.25 = 0.15 V

Now that we have all the data, we can solve for the [Ni²⁺]:

0.133 = 0.15 - 0.05916/2 log(5/[Ni²⁺])

0.133 - 0.15 = -0.05916/2 log(5/[Ni²⁺])

-0.017 = -0.02958 log(5/[Ni²⁺])

-0.017/-0.02958 = log(5/[Ni²⁺])

0.5747 = log(5/[Ni²⁺])

10^(0.5747) = 5/[Ni²⁺]

[Ni²⁺] = 5/3.7558

<h2>[Ni²⁺] = 1.33 M</h2>

7 0

3 years ago

I need help please ASAP

Yuliya22 [10]

Answer:

the name for NO is nitrogen monoxide

6 0

2 years ago

The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.

Brilliant_brown [7]

Answer : The rate of reaction is,

$Rate=4.77\times 10^{-19}M/s$

The appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)$

The rate law expression will be:

$Rate=k[NO_2][O_3]$

Given:

Rate constant = $k=1.69\times 10^{-4}M^{-1}s^{-1}$

$[NO_2]$ = $1.77\times 10^{-8}M$

$[O_3]$ = $1.59\times 10^{-7}M$

$Rate=k[NO_2][O_3]$

$Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})$

$Rate=4.77\times 10^{-19}M/s$

The expression for rate of appearance of $NO_3$ :

$\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}$

As, $\text{Rate of reaction}=4.77\times 10^{-19}M/s$

So, $\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s$

Thus, the appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

7 0

3 years ago