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Natalka [10]
3 years ago
5

What would be the pressure of a 5.00 mol sample of Cl₂ at 400.0 K in a 2.00 L container found using the van der Waals equation?

For Cl₂, a = 6.49 L²・atm/mol² and b = 0.0652 L/mol.
Chemistry
1 answer:
marusya05 [52]3 years ago
8 0

Answer:

57.478atm

Explanation:

T = 400k

n = 5mol

v = 2.00

a = 6.49L^{2}

b = 0.0652L/mol

R = 0.08206

Formula

P =  \frac{RT}{(\frac{v}{n} )-b} - \frac{a}{(\frac{v}{n} )^{2} }

P = \frac{0.08206 * 400}{(\frac{2.00}{5.00} )-0.0652} - \frac{6.49}{(\frac{2.00}{5.00} )^{2} }

P = \frac{32.824}{0.3348} - \frac{6.49}{0.16}  

P = 98.041 - 40.563

P = 57.478atm

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How many moles of cacl2 are needed to make 0.250 liters of 2.250m solution
ANTONII [103]

Answer:

.562 mol (with respect to signifigant figures)

Explanation:

to find moles we must use concentration to cancel liters.

2.250 mol/L x .250 L = .5625 mol

8 0
3 years ago
A student is quickly pipetting 129ul water and observes many bubbles in the tip. You as the second student demonstrate the corre
KonstantinChe [14]
Push the pushbutton to the first stop before inserting the tip into the solution, and then put the tip in the solution tube.
6 0
4 years ago
What is the temperature 51°C expressed in kelvins?
Evgesh-ka [11]

Answer:

The answer is

<h2>324 K</h2>

Explanation:

To express a given temperature in degree Celsius to Kelvin simplify add 273 to the given value

That's

<h3>K = 273 + °C</h3>

where

K = Kelvin

°C = value in degree Celsius

From the question the value to be converted is 51°C

It's equivalent in Kelvin is

K = 273 + 51

We the final answer as

<h2>324 K</h2>

Hope this helps you

4 0
3 years ago
A = ε l c. (a) Define the terms in the formula: A = ε l c. (Pick your answers using the letter of the correct definition.)
VladimirAG [237]

Explanation:

Using Beer-Lambert's law :

Formula used :

A=\epsilon \times c\times l

where,

A = absorbance of solution

c = concentration of solution

l = length of the cell

\epsilon = molar absorptivity of this solution

According to question:

A = (C) : absorbance measured by the spectrometer

c = (B) : concentration, in mol/L, of the stock solution from which the sample was made

l = (A): pathlength of light through the cell

ε =  (D) : molar absorptivity, a constant unique to that substance at that wavelength

7 0
3 years ago
Cerium (IV) ions are strong oxidizing agents in acid ic solution, oxidizing arsenio us acid to arsen ic acid accor ding to the f
kirill [66]

Answer:

0.2042 M is the original concentration of Ce^{4+} (aq) in the titrating solution.

Explanation:

Mass of As_2O_3 = 0.217 g

Moles of As_2O_3=\frac{0.217 g}{198 g/mol}=0.001096 mol

1 mole of As_2O_3 have 2 mole of As  and 1 mole of H_3AsO_3 have 1 mole of As.

So, from 1 mole of As_2O_3 we will have 2 moles of H_3AsO_3

Then from 0.001096 mol of As_2O_3 :

2\times 0.001096 mol=0.002192 mol of H_3AsO_3

2Ce^{4+}(aq)+H_3AsO_3(aq)+3H_2O(l)\rightarrow 2Ce^{3+}(aq)+H_3AsO_4(aq)+2H^+(aq)

According to reaction, 1 mole of H_3AsO_3 reacts with 2 mole of cerium (IV) ions,then 0.002192 mol of

\frac{2}{1}\times 0.002192 mol=0.004384 mol of cerium (IV) ions.

Volume of the  acidic cerium{IV) sulfate = 21.47 ml =0.02147 L

1 mL = 0.001 L

concentration = \frac{Moles}{Volume(L)}

[Ce^{4+}]=\frac{0.004384 mol}{0.02147 L}=0.2042 M

0.2042 M is the original concentration of Ce^{4+} (aq) in the titrating solution.

8 0
3 years ago
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