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3 years ago

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The first-order rate constant for the decomposition of n2o5, 2n2o5(g)→4no2(g)+o2(g) at 70∘c is 6.82×10−3 s−1. suppose we start w

ith 2.00×10−2 mol of n2o5(g) in a volume of 2.3 l . you may want to reference (page) section 14.4 while completing this problem. part a how many moles of n2o5 will remain after 4.0 min ?,

1 answer:

torisob [31]3 years ago

7 0

The rate constant for 1st order reaction is

K = (2.303 /t) log (A0 /A)

Where, k is rate constant

t is time in sec

A0 is initial concentration

(6.82 * 10-3) * 240 = log (0.02 /A)

1.63 = log (0.02 /A)

-1.69 – log A = 1.63

Log A = - 0.069

A = 0.82

Hence, 0.82 mol of A remain after 4 minutes.

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Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

2 years ago

Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 264 mL .

gavmur [86]

The formula for density is:

D = m/v

We can use the formula to figure out the mass because we already know two of the three values (we are given the density and volume), so we only have to solve for <em>m. </em>If we plug our given values into the formula, we get:

2.70 = m / 264

Now, all we need to do is solve for <em>m</em>. The goal is to get <em>m</em> on one side of the equation, and all we have to do is multiply each side of the equation by 264:

264 × 2.70 = (m÷264) × 264

264 × 2.70 = m

m = 712.8

The mass of the piece of aluminum is 712.8 grams.

4 0

2 years ago

Which substance below will exhibit hydrogen bonding between the molecules of the substance?

Morgarella [4.7K]

I think it might just might be e

5 0

3 years ago

You are asked to construct a segment bisector by paper folding. which is the best description your process? a use a ruler to mea

Orlov [11]

The best description of the process to draw the segment is B. b place a compass at one endpoint of the segment and trace an arc. do the same from the other endpoint. draw a line through the intersections of the arcs.

<h3>What is a segment bisector?</h3>

It should be noted that a segment bisector simply means a line to segment which cuts another line segment at the center but dividing the lines into two equal halves.

In this case, based on the information given, the best description of the process to draw the segment is to place a compass at one endpoint of the segment and trace an arc, do the same from the other endpoint. draw a line through the intersections of the arcs.

In conclusion, the correct option is B.

Learn more about segment on:

brainly.com/question/2437195

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6 0

1 year ago

A student with an unknown code of Q3A conducts a qualitative analysis of her unknown. Her results indicated the formation of a y

insens350 [35]

Answer:

Grey precipitate implies the presence of silver ions

Yellow precipitate implies the presence of lead II ions

Explanation:

Qualitative analysis provides us a quick method of identifying ions present in a sample by chemical reactions involving simple reagents. Precipitates having a unique colour is formed. The identity of ions in the sample is deduced from the colour of precipitate obtained when particular reagents are added.

In the question, a precipitate containing silver ions upon standing turn into grey colour. Similarly, lead II ions give a yellow precipitate.

6 0

2 years ago