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juin [17]
4 years ago
6

Given the following reaction: 3D(g) + E(g) + 2F(g) → 5G(g) + 4H(g)

Chemistry
1 answer:
evablogger [386]4 years ago
4 0

Answer:

Explanation:

Hello,

The law of mass action, allows us to know the required amounts, thus, for this chemical reaction it is:

\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}

Now, we answer:

(a)

\frac{d[H]}{dt}=4*\frac{1}{-3} *(-0.12M/s)=0.16M/s

(b)

\frac{d[E]}{dt}=-1*\frac{1}{5} *(0.2M/s)=-0.04M/s

(c) Since no initial data is specified, we could establish the rate of the reaction as based of the law of mass action:

r=\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}

Thus, any of the available expressions are suitable to quantify the rate of the reaction.

Best regards.

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3 years ago
How many moles are in 15.2 grams of Calcium (Ca)?
nalin [4]

Answer:

There are 0.379 moles of Ca in the given sample.

Explanation:

Hey there!

We're given that we have 15.2 grams of Ca (Calcium). We need to convert grams to moles.

There are many different conversion types for moles. A few examples are:

  • grams to moles
  • moles to grams
  • grams to particles
  • moles to liters

There are really endless possibilities for conversion factors. Since we want to find the number of moles in a certain sample, we first need to define what a mole is.

  • <u>Mole</u> - a unit of measurement in chemistry

A mole is one unit of a certain item. For instance, one cup would be equivalent to one mole. Moles are simply a hypothetical unit in chemistry that act as a placeholder.

<u>Now, how do we find the number of moles?</u>

We know that a mole of something is one of that substance. In this case, we're referring to one atom of Calcium.

Therefore, we need to reference a periodic table to understand what the mass of one atom of calcium is equivalent to. I'll be using the official AP Chemistry periodic table as issued by the College Board.

When we reference the periodic table, we see the mass of one atom of calcium is equivalent to 40.08 amu (atomic mass units).

We learn in chemistry that we can use the terms amu and grams interchangeably, so we have learned that one mole of calcium is equivalent to 40.08 grams.

<u>How do we work this out?</u>

Now, we need to find out how many moles are in 15.2 grams of calcium. We can use dimensional analysis to work this out.

Our starting unit will be 15.2 grams of Ca.

  • \text{15.2 grams Ca}

We want to convert this into moles, so we need to cancel out our grams value. Additionally, since we know that we have 40.08 grams of Ca in one mole, we can set these up as two ratios:

  • \displaystyle \frac{\text{40.08 grams Ca}}{\text{1 mole Ca}}
  • \displaystyle \frac{\text{1 mole Ca}}{\text{40.08 grams Ca}}

These ratios can be used to express the relationship between the moles of calcium and the mass of one calcium atom. However, we need our value of grams to cancel out, so since we will be using cross multiplication in our dimensional analysis, we need to use the ratio with grams on the bottom.

Now that we have selected our ratio, we can set up a multiplication problem of our two ratios:

\displaystyle \text{15.2 grams Ca} \times \frac{\text{1 mole Ca}}{\text{40.08 grams Ca}}

We assume that we have a denominator of 1 below 15.2 grams Ca.

Now, we perform our multiplication.

\displaystyle \frac{\text{15.2 grams Ca} \ \times \ \text{1 mole Ca}}{\text{40.08 grams Ca}}

We see that our remaining unit is going to be mole(s) Ca, so we can go ahead and drop our grams Ca unit.

\displaystyle \frac{\text{15.2} \ \times \ \text{1 mole Ca}}{\text{40.08}}

Now, we need to simplify a little bit.

\displaystyle \frac{\text{15.2}}{\text{40.08}} = \text{? moles Ca}

Finally, let's divide our fraction and evaluate our equation to find the value of moles Ca.

\displaystyle \frac{15.2}{40.08} = \text{0.379241517 moles Ca}

<u>Are we done yet?</u>

Not just yet! To complete our work, we need to find the number of significant figures we can use. We are given two values in which we can use to judge how many significant figures we are limited to.

  • 15.2 -- Three significant figures
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In chemistry, we always use the least number of significant figures for precision calculations, so we need to round our answer to three significant figures.

0.379241517 \approx 0.379

Therefore, there are approximately 0.379 moles of Ca in the given sample.

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LuckyWell [14K]
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when ΔG° = - R*T*㏑K

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T is the temperature in Kelvin = 25+273 = 298 K

and K is the equilibrium constant = 4.5 x 10^-4

so by substitution:

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