All categories

3 years ago

1. Explain the two processes that keep atmospheric oxygen and carbon dioxide at stable levels?

1 answer:

4 0

Cellular Respiration and Photosynthesis. Photosynthesis is the process of when plants use sunlight to make foods from carbon dioxide and water, to later on make oxygen. Cellular respiration is the process through which cells convert sugars into energy.

You might be interested in

Will give brainliest and extra points please help ASAP

Viefleur [7K]

The Correct answer to this question is translation

3 0

3 years ago

Read 2 more answers

Salicylamide can undergo an iodination by electrophilic aromatic substitution. Arrange the procedural steps in order to iodinate

Tanzania [10]

The procedural steps in order to get iodinate salicylamide are as follows:

It begins first at a laboratory.
Secondly, one has to dissolve salicylamide in ethanol.
Then one has to add sodium iodide.
Later on, one has to add sodium hypochlorite to the ice cold solution of salicylamide and sodium iodide.
Thereafter one has to sodium thiosulfate.
It is better to Acidify by a person adding about 10% of HCl.
The one has to collect the crude product through the use of vacuum filtration.
Lastly one then Recrystallize from hot ethanol.

<h3>What is Electrophilic aromatic substitution?</h3>

The Electrophilic aromatic substitution is known to be a kind of an organic reaction where an atom is said to be added or attached to a kind of aromatic structure (that is made up of hydrogen) is said to be replaced by an electrophile.

Learn more about electrophilic aromatic substitution from

brainly.com/question/14908357

4 0

2 years ago

A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = $-\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B$

If $C_{A0}$ and $C_{B0}$ are the inital concentrations and x the concentration reacted at time t, so $C_A=C_{A0} -x$ and $C_B=C_{B0} -x$ and the rate at time t is written as:

$-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)$

Separating variables and integrating

$\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt$

The integral in left side is solved by partial fractions, it can be used integral tables

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt$

Using logarithm properties (ln x - ln y = ln(x/y))

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a $C_{B0}$ of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, $C_{A0}=0.075$ , $C_{B0}=0.05$ , $C_{B}=0.02$ , it implies that the quantity reacted, x, is 0.03 and $C_{A}=0.075-x = 0.045$ . Then, the value of k would be

$kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})$

$k = 16.21 \frac{dm^3}{mol*1h}$

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

but suppossing that $C_A= C_{A0}/2=0.0375$ so $C_B=C_{B0}- C_{A0}/2=0.0125$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})$

$t=1.71 h = 1.71*3600 s = 6.1*10^3 s$

For reactant B, $C_B= C_{B0}/2=0.025$ so $C_A=C_{A0}- C_{B0}/2=0.05$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})$

$t=0.71 h = 0.7*3600 s = 2.5*10^3 s$

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0

2 years ago

Which action must occur for an enzyme to catalyze a chemical reaction?

Butoxors [25]

Answer:

correct substrate to bind the active site of the enzyme.

Explanation:

For an enzyme to catalyze a chemical reaction it is very essential for the correct substrate to bind the active site of the enzyme.

The Active site consists of two parts

1.Binding site

2. Catalytic site

The binding site consists of amino acid residues that bind to the correct substrate while the catalytic site has the amino acids that lead to the catalysis.

The active site's shape is such that only the right substratum easily binds to it and thus the catalytic reaction occurs.

4 0

2 years ago

A science teacher demonstrated physical and chemical changes by using a fork to hold a marshmallow over a lit candle

arlik [135]

Physical and chemical changes are the property of the reactions. The best observation evident of the chemical change is the deposition of the brown crust over the marshmallow.

<h3>What is a chemical change?</h3>

A chemical change is a change that results in the formation of a new substance that cannot be changed back into its original form. They are irreversible and include decomposition, combination, and addition reactions.

The color change of the marshmallow from white to brown shows the chemical change that cannot be reversed back to the original form while the melting of the candle and heating of the metallic fork are physical changes.

Therefore, the burning of marshmallows is a chemical change.

Learn more about chemical change here:

brainly.com/question/2268111

#SPJ4

3 0

1 year ago