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3 years ago

Why are coastal zones productive?

1 answer:

7 0

The surface waters in regions of coastal up-welling are cold and nutrient-rich, promoting robust growth of plants and the animals that feed on them. ... When too many nutrients are released into seas, bays, and estuaries, it can create an overabundance of decaying plants and animals, depleting oxygen from the water.

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Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

$5.4\times 10^{-3}=k(0.30)^a(0.050)^b$ ....(1)

Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

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3 years ago

Word Bank: automobiles, fossil fuels, plastic packaging, planes, trains _________________________ requires oil to create it.

Ymorist [56]

Plastic packaging requires oil to create it

7 0

3 years ago

How do you actually know the chemical formulae for the ionic compound

qaws [65]

If you have to write the chemical formula of a simple, binary ionic compound given the name of the compound, you follow a set of three steps. Let's go through them using magnesium chloride as an example. Write the symbols for the cation and the anion: Mg and Cl. Determine the charge on the cation and anion.

3 0

3 years ago

Miss White wants to make a (decidedly not poisoned) apple pie. The recipe she is

Rina8888 [55]

So you have to multiply 786,3 times 0.98 to get the mass the mass is 770.57 if u round its 770.6 or 771

5 0

3 years ago

Design an experiment to collect data that supports the claim that a 1.0M NaCl solution is a homogenous mixture.

Alika [10]

The claim is that NaCl mixture is a homogeneous mixture.
Homogeneous mixture means that the components of the mixtures cannot be determined or separated by the naked eye. However, these components can be separated using physical means, such as boiling, evaporation and condensation which will be used in this experiment.

First, we need to prepare one molar solution of NaCl. To do so, we will dilute a mass of 58.44 grams (molar mass of NaCl) in 1 liter of water.
By this, we will have NaCl solution.

We can notice that once the NaCl is diluted in water, all what you can see is a clear solution. You cannot see the separate particles of NaCl in water.
..............> observation I

Now, we will heat this solution until it boils and water starts evaporating. We will place a cold surface above the steam coming out from the boiling solution.

What we will observe is that when all the water evaporates, we can see white precipitate of NaCl in the bottom of the container. Examining the cold surface placed above the steam, we can see that the water has condensed on this surface.
.........>observation II

Based on this, we managed to use boiling, evaporation and condensation (physical methods) to restore the components of the solution separately.
.............>conclusion

Based on observation I, observation II and the conclusion. we were able to prove that NaCl solution is a homogeneous mixture.

7 0

3 years ago