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Xelga [282]
3 years ago
11

Is it true or false that some ecosystems can recover after a disturbance

Chemistry
2 answers:
Contact [7]3 years ago
7 0

Answer: i think  no but what kind of distubance

Explanation:

Sergio039 [100]3 years ago
3 0

Answer is True. They can in one lifetime by either humans or natural

You might be interested in
What is the pressure of a mixture of oxygen, nitrogen and carbon dioxide gases if the pressures of these gases are as follows: P
AleksandrR [38]

Answer:

his is an example of a first-year chemistry question where you must first convert two of the pressures to the units of the third and add them up, per Dalton’s law of additive pressures. There are three possible answers, one for each of the three pressure units.

1 atm = 760 torr …… torr and mm Hg are the same

1 atm = 101.3 kPa

Dalton’s law:

P(total) = P(O2) + P(N2) + P(CO2)

Explanation:

Gases will assume whatever pressure depending on the equation of state of the mixture (in this case) and the volume htey are contained in. That could be the ideal gas law and simple mixing law, If you are quoting the partial pressures which you call simply “the pressure” of each gas, and that these refer to their values in the present mixture, then yes, we would add them up. The pressures are low enough for the ideal gas law to apply provided the temperature is not extremely low as well .

8 0
3 years ago
Calculate the pKa of lactic acid (CH3CH(OH)COOH) given the following information. 3.005 grams of potassium lactate are added to
snow_lady [41]

Answer:

\displaystyle \text{p} K_a \approx 3.856

Explanation:

Because 3.005 grams of potassium lactate is added to 100. mL of solution, its concentration is:


\displaystyle \begin{aligned} \left[ \text{KC$_3$H_$_5$O$_3$}\right]  & = \frac{3.005\text{ g KC$_3$H_$_5$O$_3$}}{100.\text{ mL}} \cdot \frac{1\text{ mol KC$_3$H_$_5$O$_3$}}{128.17 \text{ g KC$_3$H_$_5$O$_3$}} \cdot \frac{1000\text{ mL}}{1\text{ L}} \\ \\ &= 0.234\text{ M}\end{aligned}

By solubility rules, potassium is completely soluble, so the compound will dissociate completely into potassium and lactate ions. Therefore, [KC₃H₅O₃] = [C₃H₅O₃⁺]. Note that lactate is the conjugate base of lactic acid.

Recall the Henderson-Hasselbalch equation:

\displaystyle \begin{aligned}\text{pH} = \text{p}K_a + \log \frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]} \end{aligned}

[Base] = 0.234 M and [Acid] = 0.500 M. We are given that the resulting pH is 3.526. Substitute and solve for p<em>Kₐ</em>:

\displaystyle \begin{aligned} (3.526) & = \text{p}K_a + \log \frac{(0.234)}{(0.500)} \\ \\ 3.526 & = \text{p}K_a + (-0.330) \\ \\ \text{p}K_a & = 3.856\end{aligned}

In conclusion, the p<em>Kₐ </em>value of lactic acid is about 3.856.

5 0
2 years ago
What was Thomson's model of the atom called?
trasher [3.6K]
The answer is C: The plum-pudding model
3 0
3 years ago
A weak acid, HF, is in solution with dissolved sodium fluoride, NaF. If HCl is added, which ion will react with the extra hydrog
kumpel [21]

Answer:

c. F1-

Explanation:

In this chemical reaction the expression is:

HF  +  NaF   →  NaHF2

The ion that always keep the negative charge is the fluorine with a -1, if in this mixture there is more positive ions (H1+) the negative ion (F1-) will join with them.

Remember that also the Cl1- will be free, but the fluorine is more reactive than the fluorine.

5 0
3 years ago
Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to
zhenek [66]

Answer:

<u>D. It will decrease by a factor of 4</u>

Explanation:

According to the question , the equation follows :

A+B\rightarrow C+D

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

Rate\ \alpha [A]^{a}[B]^{b}

r=[A]^{a}[B]^{b}.................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

r'=[A']^{a}[B']^{b}

Put the value of [A'] , [B'] and r' in the above equation:

2r=[2A]^{a}[B]^{b}...........(2)

Divide equation (1) by (2) we , get

\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}

2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}

Here A and A cancel each other

B and B cancel each other

We get,

2= 2^{a}\times 1^{b}

1^b = 1 ( power of 1 = 1)

2= 2^{a}

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1  = 2

b = 2

Hence the rate law becomes :

r=[A]^{a}[B]^{b}

<u>r=[A]^{1}[B]^{2}.............(3)</u>

Look in the question now, it is asked to calculate the concentration of [B],if  cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

r'=[A]^{1}[B']^{2}

r'=[A]^{1}(\frac{1}{2}[B])^{2}

r'=\frac{1}{4}[A]^{1}[B]^{2}............(a)

But

r=[A]^{a}[B]^{b}..............(b)

Compare equation (a) and (b) , we get

new rate r' =

<u>r' = 1/4 r</u>

7 0
3 years ago
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