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4 years ago

HELP FAST 20 POINTS

Chemistry

2 answers:

schepotkina [342]4 years ago

8 0

D) Transportation is the correct answer

Naddik [55]4 years ago

7 0

The answer is D because it’s the movement of food substance from the stems to growing tissues and storage tissues,
The answer A is incorrect because it talks about the cycle

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Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

8 0

3 years ago

What is the pOH of a solution with [OH-] = 1.4 x 10-13?

lukranit [14]

Answer:

C. 12.85

Explanation:

just did

4 0

3 years ago

Under which conditions (in terms of kinetic energy) does boiling occur? How is the kinetic energy in the liquid increased?

Simora [160]

Explanation:

Boiling is defined as a process in which vapor pressure of a liquid substance becomes equal to the atmospheric pressure.

During this change liquid and vapors remain in equilibrium and the equation for this change is as follows.

$X(l) \rightleftharpoons X(g)$

Therefore, when boiling takes place then average kinetic energy of particles in liquid phase equals to the average kinetic energy of particles in vapor phase.

Hence, we can increase the kinetic energy of particles in liquid phase by increasing the temperature because kinetic energy is directly proportional to temperature as follows.

K.E = $\frac{3}{2}kT$