All categories

2 years ago

Describe a Sankey Diagram and how it can be used.

Chemistry

1 answer:

jolli1 [7]2 years ago

4 0

Explanation:

Sankey diagrams , which are typically used to visualize energy transfers between processes, are named after the Irishman Matthew H. P. R. Sankey, who used this type of diagram in a publication on energy efficiency of a steam engine in 1898.

Sankey diagrams are ideal for visually representing energy balances.

how to use

1.Overview. The Sankey diagram displays how quantities are distributed among items between two or more stages.

2.Add a Sankey diagram. Choose the Data Visualization or Re-Visualize option from the toolbar and select Sankey Diagram.

3.Change link color and width.

4.Change node color.

5.Change labels and tooltips.

You might be interested in

In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti

Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

$\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }$ ⇄ $\mathtt{3SO_{2(l)}}$

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

$K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}$

However, since we are dealing with liquids solutions;

$K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}$ since the activity of $a_{so_3}$ is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

$K = \dfrac{1}{Pso_2Po_2^{1/2}}$

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

$\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\ \simeq -68 \ kJ/mol$

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

$\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}$

$\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K } \\ \\ K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\ K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }$

$K =7.98390356\times 10^{11} \\ \\ \mathbf{K = 7.98 \times 10^{11}}$

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

$Q= \dfrac{1}{Pso_2Po_2^{1/2}}$

Since we are dealing with liquids;

$Q= \dfrac{1}{1 \times 1^{1/2}}$

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

8 0

3 years ago

A 2.87 g sample of carbon reacts with hydrogen to form 3.41 g of car fuel. What is the empirical formula for the car fuel?

olga_2 [115]

CH₇ is the empirical formula of the car fuel.

Explanation:

To find the empirical formula we use the following algorithm.

First divide each mass the the molar weight of each element:

for carbon 2.87 / 12 = 0.239

for hydrogen 3.41 / 2 = 1.705

And now divide each quantity by the lowest number which is 0.239:

for carbon 0.239 / 0.239 = 1

for hydrogen 1.705 / 0.239 = 7.13 ≈ 7

The empirical formula of the car fuel is CH₇.

I have to tell you that in reality this formula is wrong because is not possible to exist. However the algorithm for finding the empirical formula is right, the problem may reside in the amounts of carbon and hydrogen given.

Learn more about:

empirical formula

brainly.com/question/5297213

#learnwithBrainly

5 0

3 years ago

Please help, with step by step work

natali 33 [55]

$\qquad$ ☀️ $\pink{\bf{ {Answer = \: \: 85.57g }}}$

Molar mass of $\bf Cu_2O$

$\qquad$ $\twoheadrightarrow\sf 63.546 \times 2 +16$

$\qquad$ $\pink{\twoheadrightarrow\bf 143.092 g}$

<u>As we know</u>–

1 mol = $\bf 6.02×10^{23}$ formula units

1 mol $\bf Cu_2O$ = 143.092 g = $\bf 6.02×10^{23}$ formula units

Henceforth –

$\bf 3.60×10^{23}$ formula units $\bf Cu_2O$ –

$\qquad$ $\sf :\implies \dfrac{143.092 \times3.60×10^{23 }}{6.02×10^{23}}$

$\qquad$ $\sf :\implies \dfrac{143.092 \times3.60×\cancel{10^{23 }}}{6.02×\cancel{10^{23}}}$

$\qquad$ $\pink{:\implies\bf 85.57 g}$

5 0

2 years ago

Why is sterilising the water rarely used in treating the sewage water?

Marina86 [1]

Answer:

Sterilizing the water rarely used in treating the sewage water is described below in details.

Explanation:

Sterilization is the application of a chemical or physical system to eradicate all microorganisms, concluding comprehensive amounts of bacterial content. Heat treatment is the most relevant approach for the sterilization of waste-water that accommodates large levels of masses. well now sterilizing the water is un-usually practiced in managing the sewage water.

7 0

3 years ago

Question 4 options: what is the volume of 1.2 moles of water vapor at stp?

Law Incorporation [45]

What is the volume of 1.2 moles of water vapor at STP?
The answer is 26.9

3 0

3 years ago