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2 years ago

4 points

Chemistry

2 answers:

kogti [31]2 years ago

6 0

Explanation:

groups are the vertical rows of element and each group have the same no of valence electron

periods are the horizontal row of element on the basis of no of orbital

Ilya [14]2 years ago

4 0

The vertical columns are groups

The horizontal rows are periods

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Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C

ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)

The enthalpy of reaction (1) is given by:

$\Delta H = \Delta H_{r} - \Delta H_{p}$ (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond

7O₂: 7 moles of 1 O=O bond

4CO₂: 4 moles of 2 C=O bonds

6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

$\Delta H = \Delta H_{r} - \Delta H_{p}$

$\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})$

$\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})$

$\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol$

$\Delta H = -2860 kJ/mol$

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

I hope it helps you!

7 0

2 years ago

Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the fol

zvonat [6]

Answer:

1.58x10⁻⁵

2.51x10⁻⁸

0.0126

63.10

Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = $10^{-4.8}$

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

2.1 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -7.6

[In-]/[HIn] = $10^{-7.6}$

[In-]/[HIn] = 2.51x10⁻⁸

iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

[In-]/[HIn] = $10^{-1.9}$

[In-]/[HIn] = 0.0126

iv) pH = 11.5

11.5 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = 1.8

[In-]/[HIn] = $10^{1.8}$

[In-]/[HIn] = 63.10

6 0

3 years ago

In carbon dioxide (CO2), there are two oxygen atoms for each carbon atom. Each oxygen atom forms a double bond with carbon, so t

katrin2010 [14]

Answer: D) Eight

Just pretend this part doesn't exist nope nothing to see here the answer is correct on edge2020 I swear on my brainly points

4 0

3 years ago

Read 2 more answers

Which of the following molecules experience dipole-dipole forces as its strongest IMF? A) H2 B) SO2 C) NH3 D) CF4 E) BCl3

Xelga [282]

Answer: $NH_3$

Explanation:

a) $H_2$ : This is a non polar covalent compound which are held by weak vanderwaal forces of attraction.

b) $SO_2$ : This is a covalent compound which is polar due to the presence of lone pair of electrons and are held by dipole-dipole forces of attraction.

c) $NH_3$ : These are joined by a special type of dipole dipole attraction called as hydrogen bond. It forms between electronegative nitrogen atom and hydrogen atom and is the strongest interaction.

d) $CF_4$ : This is a covalent compound and is non polar which are held by weak vanderwaal forces of attraction.

e) $BCl_3$ : This is a covalent compound and is non polar which are held by weak vander waal forces of attraction.

5 0

2 years ago

What makes metals, in particular, good conductors of electricity?

hichkok12 [17]

Answer:

the answer is A(the ability of electrons to flow through out the metal)

3 0

2 years ago