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3 years ago

Evidence for sea-floor spreading has come from

Chemistry

2 answers:

AVprozaik [17]3 years ago

8 0

Answer:

your answer is B.

Explanation:

have a nice day/night!!

Taya2010 [7]3 years ago

6 0

Answer B because of how it differs throughout time

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Even at high T, the formation of NO is not favored:

Readme [11.4K]

Answer:

3,16x10⁻³M

Explanation:

For the reaction:

N₂(g) + O₂(g) ⇄ 2NO(g). The kc is defined as:

kc = [NO]² / [N₂] [O₂] = 4,10x10⁻⁴ (1)

If you add in a 1,0L container 0,25 mol of N₂ and 0,10 mol of O₂, concentrations in equilibrium will be:

[N₂] = 0,25M - x

[O₂] = 0,10M - x

[NO] = 2x

Replacing in (1):

[2X]² / [0,25-x] [0,10-x] = 4,10x10⁻⁴

[2X]² / 0,025 - 0,35x + x²= 4,10x10⁻⁴

4X² = 4,10x10⁻⁴x² - 1,435x10⁻⁴x + 1,025x10⁻⁵

3,99959x² + 1,435x10⁻⁴x - 1,025x10⁻⁵ = 0

Solving for x:

x = -0,0016 (Wrong answer, there is no negative concentrations)

x = 0,00158 (Right answer)

As molar concentration of NO in equilibrium is 2x:

[NO] = 2x = 2×0,00158 = 3,16x10⁻³M

I hope it helps!

6 0

4 years ago

The density of magnesium is 1.7/cm3, and the density of iron is 7.9g/cm3. Consider a block of iron with a mass of 819 g. What is

zepelin [54]

The density of magnesium is 1.7/cm3, and the density of iron is 7.9g/cm3. Consider a block of iron with a mass of 819 g. What is the mass of a block of magnesium that has the same volume as the block of iron?

3 0

3 years ago

Please help fast thanks

timurjin [86]

Answer:

I do not know if I am correct but I think it is the letter D I hope it helps you if it is correct pls brainliests

7 0

3 years ago

Why thre is no atmosphere in moom,?

goblinko [34]

The gravitational pull of the moon is not strong enough to attract a significant atmosphere.

3 0

3 years ago

A chemistry student needs 10.0g of ethanolamine for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the

iogann1982 [59]

Answer: The volume of ethanolamine the student should pour out is $9.80cm^3$

Explanation:

To calculate the volume of ethanolamine we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanolamine = $1.02g/cm^3$

Mass of ethanolamine = 10.0 g

Volume of ethanolamine = ?

Putting values in above equation, we get:

$1.02g/cm^3=\frac{10.0g}{\text{Volume of ethanolamine}}\\\\{\text{Volume of ethanolamine}=\frac{10.0g}{1.02g/cm^3}=9.80cm^3$

Thus the volume of ethanolamine the student should pour out is $9.80cm^3$

4 0

4 years ago