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3 years ago

The organ system that moves air into, around, and out the body

Chemistry

1 answer:

Umnica [9.8K]3 years ago

6 0

Its a Respiratory System, hope that helped your question :)

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A) For the following reaction, K > 1. Classify each of the reactants and products based on their strength as Bronsted-Lowry a

Aleks04 [339]

Answer:

Explanation:

According to Bronsted-Lowry acids or base theory , the reagent capable of giving hydrogen ion or proton will be acid and that which accepts hydrogen ion or proton will be base .

C₉H₇N + HNO₂ ⇄ C₉H₇NH⁺ + NO₂⁻

If K > 1 , reaction is proceeding from left to right .

Hence HNO₂ is giving H⁺ or proton and C₉H₇N is accepting proton to form

C₉H₇NH⁺ .

Hence HNO₂ is bronsted acid and C₉H₇N is bronsted base .

B )

when K < 1 , reaction above proceeds from right to left . That means

C₉H₇NH⁺ is giving H⁺ so it is a bronsted acid and NO₂⁻ is accepting H⁺ so it is a bronsted base .

Hence , NO₂⁻ is a bronsted base and C₉H₇NH⁺ is a bronsted acid .

4 0

3 years ago

Opening a soda bottle illustrates how a decrease in pressure will___the amount of gas that can be dissolved in water.

strojnjashka [21]

C. Increase :)
I would choose C

3 0

3 years ago

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How does the corona of the Sun affect earth

SIZIF [17.4K]

Answer:

I don't think it does affect Earth

Explanation:

The sun is too far away to get corona let alone spread it.

6 0

3 years ago

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Determine the equilibrium constant, Keq, at 25°C for the reaction

adelina 88 [10]

Explanation:

The given chemical reaction is:

$2Br^- (aq) + I_2(s) Br_2(l) + 2I^- (aq)$

$E^ocell=oxidation potential of anode + reduction potential of cathode\\$

The relation between Eo cell and Keq is shown below:

$deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell$

The value of Eo cell is:

Br- undergoes oxidation and I2 undergoes reduction.

Reduction takes place at cathode.

Oxidation takes place at anode.

Hence,

$E^ocell= (-1.07+0.53)V\\=-0.54V$

F=96485 C/mol

n=2 mol

R=8.314 J.K-1.mol-1

T=298K

Substitute all these values in the above formula:

$ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3$

Answer:

Keq=6.13x10^33

3 0

3 years ago

If hydrochloric acid has a [H+] of 1.2 x 10-2 M, what is the pH?<br> 1.9<br> 0 1<br> 1.2<br> O<br> 8

Vanyuwa [196]

Answer:

Explanation:

The pH of a solution can be found by using the formula

$pH = - log [ {H}^{+} ]$

From the question we have

$pH = - log(1.2 \times {10}^{ - 2} ) \\ = 1.920818...$

We have the final answer as

Hope this helps you

7 0

3 years ago

Read 2 more answers