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expeople1 [14]
3 years ago
15

When each of the following processes reach equilibrium, does the system in question contain mostly reactants, mostly products, o

r fairly equal concentrations of both reactants and products? Justify your answers.
a. 2 H2(g) +02(g)2 H20(g)
b. H2(g + Br2(g)2HBr(g) K. 91 x 1080 at 25°C K-114 x 1021 at 25°C
Chemistry
1 answer:
jasenka [17]3 years ago
7 0

Answer:

Mostly products

Explanation:

Given the very large equilibrium constants for the two reactions we can consider them to be essentially 100 % complete to the product side.

The equilibrium constant for the reactions are given by:

a. K = p H₂O ² / (p H₂ ² x  p O₂ ) where p are the pressures  K = 91 x 10⁸⁰

b. K = p HBr ² / ( p H₂ x p Br₂ )   K = 114 x 10²¹

These two number are so big that  essentially very, very little amount of reactants are present after reaction.

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(3) 6 electrons are shared.

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4 years ago
An amphoteric species is neither an acid nor a base. True or False
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Answer: False

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8 0
3 years ago
What is the mass in grams of 5.00moles of CH4?
anastassius [24]

Answer:

1. 80g

2. 1.188mole

Explanation:

1. We'll begin by obtaining the molar mass of CH4. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Number of mole of CH4 from the question = 5 moles

Mass of CH4 =?

Mass = number of mole x molar Mass

Mass of CH4 = 5 x 16

Mass of CH4 = 80g

2. Mass of O2 from the question = 38g

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Number of mole O2 =?

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6 0
3 years ago
Astroturf is a durable artificial surface used to cover athletic fields. A soccer field 0.06214- mile-long by 253 ft wide is cov
love history [14]

Answer:

The weight of the Astroturf is 179,684.31 Newtons.

Explanation:

Length of a soccer field = 0.06214 mile = 328.0992 feet

(1 mile = 5280 feet)

Breadth of a soccer field  = 253 feet

Length of a Astroturf which soccer field is to be covered, l = 328.0992 feet

Breadth of a Astroturf which soccer field is to be covered ,b = 253 feet

Thickness of a Astroturf with which soccer field is to be covered = h

h = ½ inch = 0.5 inch = 0.041665 feet

(1 inches = 0.08333 feet)

Volume of the Astroturf ,V= l × b × h

V=328.0992 ft\times 253 ft\times 0.041665 ft=3,458.574 ft^3

Mass of the Astroturf = m

Density of the Astroturf = d = 187 oz/ft^3

d=\frac{m}{V}

m=d\times V= 187 oz/ft^3\times 3,458.574 ft^3=646,753.35 oz

1 oz = 0.0283495 kg

m=646,743.35 oz=646,743.35\times 0.0283495 kg=18,335.13 kg

Weight of the Astroturf = W

W = mg

=W=18,335.13 kg\times 9.8 m/s^2=179,684.31 N

The weight of the Astroturf is 179,684.31 Newtons.

8 0
3 years ago
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