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2 years ago

PLEASE HELP

Chemistry

2 answers:

kondaur [170]2 years ago

8 0

Answer:

convection

Explanation:

..................

Elza [17]2 years ago

6 0

Answer:

convection

Explanation:

because hot water rise cold water sinks

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Is pentacosane a molecular compound?

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Yes

Explanation:

A molecular compound is that one which is composed of molecules formed by atoms of two or different elements. A pentacosane is an alkaline made of unbranched chain of 25 carbons atoms. It has applications in plant metabolite. The molecular formula of pentacosane is C₂₅H₅₂

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Palladium (Pd; Z 46) is diamagnetic. Draw partial orbital diagrams to show which of the following electron configurations is con

nexus9112 [7]

Answer : The electron configurations consistent with this fact is, (b) [Kr] 4d¹⁰

Explanation :

Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom are determined by the electronic configuration.

Paramagnetic compounds : They have unpaired electrons.

Diamagnetic compounds : They have no unpaired electrons that means all are paired.

The given electron configurations of Palladium are:

(a) [Kr] 5s²4d⁸

In this, there are 2 electrons in 's' orbital and 8 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital are paired but 'd' orbital are not paired. So, this configuration shows paramagnetic.

(b) [Kr] 4d¹⁰

In this, there are 10 electrons in 'd' orbital. From the partial orbital diagrams we conclude that electrons in 'd' orbital are paired. So, this configuration shows diamagnetic.

(c) [Kr] 5s¹4d⁹

In this, there are 1 electron in 's' orbital and 9 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital and 'd' orbital are not paired. So, this configuration shows paramagnetic.

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3 years ago

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

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3 years ago