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Georgia [21]
3 years ago
15

Alkali metals are found in many common substances but are rarely found in pure form. Explain why this is in the case

Chemistry
1 answer:
Evgesh-ka [11]3 years ago
6 0

Alkali metals are rarely found in pure form because they are so reactive.

Alkali metals are in Group 1 of the Periodic Table.

They need to lose only one valence electron to reach a filled valence shell

Thus, they readily react wirh water and other chemicals in the earth to form compounds.

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Hydrogen bonds within liquid water are attractions between protons and hydroxide ions. are dipole-dipole attractions. are ion-in
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Consider the formation of [Ni(en)3]2+ from [Ni(H2O)6]2+. The stepwise ΔG∘ values at 298 K are ΔG∘1 for first step=−42.9 kJ⋅mol−1
timurjin [86]

Answer:

kf = 1.16 x 10¹⁸

Explanation:

Step 1: [Ni(H₂O)₆]²⁺  + 1en → [Ni(H₂O)₄(en)]²⁺  ΔG°1 = -42.9 kJmol⁻¹

Step 2: [Ni(H₂O)₄(en)]²⁺  + 1en → [Ni(H₂O)₂(en)₂]²⁺  ΔG°2 = -35.8 kJmol⁻¹

Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en →  [Ni(en)₃]²⁺  ΔG°3 = -24.3 kJmol⁻¹

________________________________________________________

Overall reaction: [Ni(H₂O)₆]²⁺  + 3en → [Ni(en)₃]²⁺  ΔG°r

ΔG°r = ΔG°1 + ΔG°2 + ΔG°3

ΔG°r = -42.9 - 35.8 - 24.3

ΔG°r = -103.0 kJmol⁻¹

ΔG°r = -RTlnKf

-103,000 Jmol⁻¹ =  - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf

kf = e ^(-103,000/-8.31x298)

kf = e ^41.59

kf = 1.16 x 10¹⁸

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3 years ago
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3 years ago
Write a balanced chemical equation, including physical state symbols, for the combustion of gaseous butane into gaseous carbon d
Phantasy [73]

Answer:1.

1.Balanced equation

C4H10 + 9 02 ==> 5H20 +4CO2

2. Volume of CO2 =596L

Explanation:

1.Combustion of alkane is the reaction of alkanes with Oxygen. And the general equation for the combustion is;

CxHy +( x+y/4) O2 ==> y/2 02 + xCO2

Where x and y are number of carbon and hydrogen atoms respectively.

For butane (C4H10)

x=4 and y=10

Therefore

C4H10 + 9 02 ==> 5H20 +4CO2

2. Mass of butane = 0.360kg

Molar mass of C4H10 = ( 12×4 + 1×10)

= 48 +10=58g/mol= 0.058kg/mol

Mole = mass/molar mass

Mole = 0.360/0.058= 6.2moles

From the stoichiometric equation

1mole of C4H10 will gives 4moles of CO2

Therefore

6.2moles of C4H10 will gives 4 moles of 24.8 moles of CO2

Using the ideal gas equation

PV=nRT

P= 1.0atm

V=?

n= 24.8mol.

R=0.08206atmL/molK

T=20+273=293

V= 24.8 × 0.08206 × 293

V= 596L

Therefore the volume of CO2 produced is 596L

8 0
3 years ago
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