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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

How many grams of NH3 can be dissolved in 50 grams of water at 50oC?

weqwewe [10]

15 grams of NH3 can be dissolved

<h3>Further explanation</h3>

Given

50 grams of water at 50°C

Required

mass of NH3

Solution

Solubility is the maximum amount of a substance that can dissolve in some solvents. Factors that affect solubility

1. Temperature:
2. Surface area:
3. Solvent type:
4. Stirring process:

We can use solubility chart (attached) to determine the solubility of NH3 at 50°C

From the graph, we can see that the solubility of NH3 in 100 g of water at 50 C is 30 g

So that the solubility in 50 grams of water is:

= 50/100 x 30

= 15 grams

5 0

3 years ago

How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl

VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

$12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}$

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

$4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}$

Part C:

According to the balanced equation 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

$13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}$

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

$7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}$

3 0

3 years ago

You need to prepare 1 L of the citric acid/citrate buffer. You have chosen to use Method 1 (see lab presentation). Calculate the

prisoha [69]

Answer:

3.11 is the pH of the buffer

Explanation:

The pH of a buffer is obtained using H-H equation:

pH = pKa + log [Conjugate base] / [Weak acid]

Where pH is the pH of the buffer, pKa = -log Ka = 3.14 for the citric buffer and [] could be taken as the moles of each species.

The citric acid,HX (Weak acid), reacts with NaOH to produce sodium citrate, NaX (weak base) and water:

HX + NaOH → H2O + NaX

That means the moles of NaOH added = Moles of sodium citrate produced

And the resulitng moles of HX = Initial moles - Moles NaOH added

Moles HX and NaX:

Moles NaOH = 0.100L * (0.65mol / L) = 0.065 moles NaOH = Moles NaX

Moles HX = 0.300L * (0.45mol / L) = 0.135 moles HX - 0.065 moles NaOH = 0.070 moles HX

Replacing in H-H equation:

pH = 3.14 + log [0.065mol] / [0.070mol]

pH = 3.11 is the pH of the buffer

8 0

3 years ago

An atom of sodium has an atomic mass of 23 and an atomic number of 11. The atom contains

babunello [35]

The element sodium has 12 neutrons, 11 electrons and 11 protons.

8 0

3 years ago

A solution of CaCl 2 in water forms a mixture that is 42.0 % calcium chloride by mass. If the total mass of the mixture is 839.6

Lerok [7]

Answer : The masses of calcium chloride and water used were, 352.6 g and 487.0 g.

Explanation :

As we are given that 42.0 % calcium chloride by mass that means 42.0 grams of calcium chloride present in 100 grams of solution.

In the solution or mixture, 42.0 % calcium chloride and 58.0 % (100-42.0=58.0) water.

Now we have to determine the mass of calcium chloride for 839.6 grams of solution.

As, 100 grams of solution contains 42.0 grams of calcium chloride

So, 839.6 grams of solution contains $\frac{42.0}{100}\times 839.6=352.6$ grams of calcium chloride

Thus, the mass of calcium chloride used is, 352.6 grams.

Now we have to determine the mass of water.

Mass of water = Mass of solution - Mass of calcium chloride

Mass of water = 839.6 g - 352.6 g

Mass of water = 487.0 g

Thus, the mass of water used is, 487.0 grams.

6 0

3 years ago