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Zepler [3.9K]
3 years ago
13

Please Help Part B to question A

Chemistry
1 answer:
anyanavicka [17]3 years ago
8 0

Answer:

Over the last century the burning of fossil fuels like coal and oil has increased the concentration of atmospheric carbon dioxide (CO2). This happens because the coal or oil burning process combines carbon with oxygen in the air to make CO2.  as u know decreasing means going down and increasing means going up.  the reason why things decrease means something went wrong or something to happen to make it go down we all try to make things increase which means we try to make things go up instead of down. a lot of reasons is y things decrease is bc during the period of time something might go wrong or happen in a bad way that makes a lot of things decrease. for example like temperature when the temperature decreases it means it goes down and it might get cold but if the temperature increases it goes up so it might be ht or warm or maybe even cool. if something ever decreases it means during the period of time something bad happened and made it go down instead of up if something ever goes wrong that means something decreased. sometimes decreasing could be a good thing but most of the time it it bad  but there are sometimes when decreasing is good but not all the time.

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6 0
3 years ago
Which of the following are evidence that chemical change has occurred
dolphi86 [110]

Answer:

gas given off

Explanation:

that's the answer

3 0
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student mixes vinegar and baking soda inside a sealed plastic bag would the reaction be the same if baking soda and vinegar were
pychu [463]
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3 0
3 years ago
The pKa of lactic acid is 3.9. A lactate buffer will be useful from pH values ________. The pKa of lactic acid is 3.9. A lactate
Vedmedyk [2.9K]

Answer:

Explanation:

The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.  

This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since  buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .

From the Henderson-Hasselbach equation we have that

pH = pKa + log [A⁻]/[HA]

thus

0.1 ≤  [A⁻]/[HA] ≤ 10

Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.

Now we are equipped to answer our question:

pH range = 3.9 +/- 1 = 2.9 through 4.9

7 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

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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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