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2 years ago

A container of hydrogen at 172 kPa was decreased to 85.0 kPa producing a new volume of 3L. What was the original volume?

Chemistry

1 answer:

Aneli [31]2 years ago

4 0

Answer:

<h2>The answer is 1.48 L</h2>

Explanation:

In order to find the original volume we use the same for Boyle's law which is

$P_1V_1 = P_2V_2$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the original volume

$V_1 = \frac{P_2V_2}{P_1} \\$

From the question

P1 = 172 kPa = 172000 Pa

P2 = 85 kPa = 85000 Pa

V2 = 3 L

We have

$V_1 = \frac{85000 \times 3}{172000} = \frac{255000}{172000} = \frac{255}{172} \\ = 1.482558139...$

We have the final answer as

Hope this helps you

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Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the

Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by $\eta$ .

The given data is as follows.

$\eta$ = 0.82, $T_{1}$ = (21 + 273) K = 294 K

$P_{1}$ = 200 kPa, $P_{2}$ = 1000 kPa

Therefore, calculate the final temperature as follows.

$\eta = \frac{T_{2} - T_{1}}{T_{2}}$

0.82 = $\frac{T_{2} - 294 K}{T_{2}}$

$T_{2}$ = 1633 K

Final temperature in degree celsius = $(1633 - 273)^{o}C$

= $1360^{o}C$

Now, we will calculate the entropy as follows.

$\Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}$

For 1 mole, $\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

It is known that for $NH_{3}$ the value of $C_{v}$ = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

$\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

= $0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}$

= 0.0346 kJ/mol

or, = 34.6 J/mol (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0

2 years ago

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

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

The Lewis structure for CO2 has a central The Lewis structure for C O 2 has a central blank atom attached to blank atoms.

kirza4 [7]

Answer:

See the explanation

Explanation:

1) The Lewis structure for $CO_2$ has a central Carbon atom attached to Oxygen atoms.

In the $CO_2$ we will have a structure: O=C=O the central atom "carbon" we will have 2 sigma bonds and 2 pi bonds, therefore, we have an Sp hybridization. For O we have 1 pi and 1 sigma bond, therefore, we have an Sp2 hybridization.

2) These atoms are held together by double bonds.

Again in the structure of $CO_2$ : O=C=O we only have double bonds.

3. Carbon dioxide has a Carbon dioxide has a Linear electron geometry.

Due to the double bonds we have to have a linear structure because in this geometry the atoms will be further apart from each other.

4. The carbon atom is Sp hybridized.

We will have for carbon 2 pi bonds, so we will have an Sp hybridization.

5. Carbon dioxide has two Carbon dioxide has two C(p) - O(p) π bonds and two C(sp) - O(Sp2) σ bonds.

(See figures)

Figure 1: Carbon hybridization

Figure 2: Oxygen hybridization

6 0

2 years ago

For each of the following compounds, state whether it is ionic or covalent, and if it is ionic, write the symbols for the ions i

kupik [55]

Answer :

Covalent compound : It is defined as the compound which is formed by the sharing of electrons between the atoms forming a compound.

The covalent compound are usually formed when two non-metals react.

Ionic compound : It is defined as the compound which is formed when electron gets transferred from one atom to another atom.

All the polyatomic ions always form an ionic compound.

Polyatomic ions : It is a charged species that composed of two or more atoms and these charged species are bonded by the covalent bond.

For the given options:

Option A: $KClO_4$

This compound is formed by the combination of potassium, $K^+$ which is a metal and $ClO_4^-$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

Option B: $Mg(C_2H_3O_2)_2$

This compound is formed by the combination of magnesium, $Mg^{2+}$ which is a metal and $C_2H_3O_2^{-}$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

Option C: $H_2S$

Hydrogen and sulfur, both are non-metals and they will form a covalent compound.

Option D: $Ag_2S$

This compound is formed by the combination of silver, $Ag^{+}$ which is a metal and sulfur, $S^{2-}$ which is a non-metal. Thus, it will form an ionic compound.

Option E: $N_2Cl_4$

Nitrogen and chlorine, both are non-metals and they will form a covalent compound.

Option F: $Co(NO_3)_2$

This compound is formed by the combination of cobalt, $Co^{2+}$ which is a metal and $NO_3^{-}$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

6 0

2 years ago

When an aldose reacts with Barfoed's reagent, what type of organic compound forms? What type of chemical is this?

Fudgin [204]

Barfoed's test is a concoction test utilized for identifying the nearness of monosaccharides. It depends on the diminishment of copper(II) acetic acid derivation to copper(I) oxide (Cu2O), which frames a block red hasten.
Barfoed's reagent comprises of a 0.33 molar arrangement of unbiased copper acetic acid derivation in 1% acidic corrosive arrangement. The reagent does not keep well and it is, thusly, fitting to make it up when it is really required. May store uncertainly as per a few MSDS's.

7 0

2 years ago

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