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2 years ago

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The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of C3H8 is 44.1 g/mol. What mass of O2, in grams, is required t

o completely react with 0.025 g C3H8

1 answer:

Ira Lisetskai [31]2 years ago

6 0

The reaction formula of this is C3H8 + 5O2 --> 3CO2 + 4H2O. The ratio of mole number of C3H8 and O2 is 1:5. 0.025g equals to 0.025/44.1=0.00057 mole. So the mass of O2 is 0.00057*5*32=0.0912 g.

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ANSWER FOR: How did the lab activities help you answer the lesson question "How do the processes of conduction, convection, and

Alenkinab [10]

<u>Answer</u>: Conduction, convection, and radiation move energy from the Sun to Earth and throughout Earth.

Without more information about the experiment itself, I would choose the above answer as correct. All the other statements are correct, however none of them relates to the earth distribution processes on Earth. The last statement does.

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2 years ago

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The incomplete table below shows selected characteristics of gas laws.

Elenna [48]

Explanation :

In the given case different law related to gas is given. The attached figure shows the required solution.

Boyle's law states that the pressure is inversely proportional to the volume of the gas i.e.

$P\propto \dfrac{1}{V}$

$PV=k$

k is a constant.

Charle's law states that the volume of directly proportional to the temperature of the gas.

$V\propto T$

$V=kT$

Combined gas law is the combination of the pressure, volume and the temperature of the gas i.e.

$\dfrac{PV}{T}=k$

Hence, this is the required solution.

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2 years ago

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Write the electron configurations for the following ions:

Ket [755]

Answer:

Co²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷

Sn²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

Zr⁴⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

Ag⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰

S²⁻ : 1s² 2s² 2p⁶ 3s² 3p⁶

Explanation:

Cobalt (Co): atomic number 27

<u>The electronic configuration of Co in ground state: </u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷

<u>The electronic configuration of Co in +2 oxidation state (Co²⁺) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷

Tin (Sn): atomic number 50

<u>The electronic configuration of Sn in ground state: </u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p²

<u>The electronic configuration of Sn in +2 oxidation state (Sn²⁺) </u>:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

Zirconium (Zr): atomic number 40

<u>The electronic configuration of Zr in ground state:</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d²

<u>The electronic configuration of Zr in +4 oxidation state (Zr⁴⁺) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

Silver (Ag): atomic number 47

<u>The electronic configuration of Ag in ground state:</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰

<u>The electronic configuration of Ag in +1 oxidation state (Ag⁺) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰

Sulphur (S): atomic number 16

<u>The electronic configuration of S in ground state:</u>

1s² 2s² 2p⁶ 3s² 3p⁴

<u>The electronic configuration of S in -2 oxidation state (S²⁻) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶

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2 years ago

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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

Arrange the following in order of increasing boiling point: RbCl, CH3Cl, CH3OH, CH4. A. CH3OH < CH4 < CH3Cl < RbCl B. R

rosijanka [135]

Answer:

E. CH₄ < CH₃Cl < CH₃OH < RbCl

Explanation:

The molecule with the stronger intermolecular forces will have the higher boiling point.

The order of strength of intermolecular forces (strongest first) is

Ion-Ion
Hydrogen bonding
Dipole-dipole
London dispersion

RbCl is a compound of a metal and a nonmetal. It is an ionic compound, so it has the highest boiling point.

CH₃Cl has a C-Cl polar covalent bond. It has dipole-dipole forces, so it has the second lowest boiling point.

CH₃OH has an O-H bond. It has hydrogen bonding, so it has the second highest boiling point.

CH₄ has nonpolar covalent C-H bonds. It has only nonpolar bonds, so the only attractive forces are London dispersion forces. It has the lowest boiling point.

Thus, the order of increasing boiling points is

CH₄ < CH₃Cl < CH₃OH < RbCl

4 0

2 years ago