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4 years ago

20 POINTS PLEASE ANSWER

Chemistry

1 answer:

Sever21 [200]4 years ago

8 0

c. It decreases when going down a group.

<u>Explanation:</u>

Ionization energy is the energy needed to remove an electron from the neutral gaseous atom.

It is represented in kJ/mol.

As we move down the group the ionization energy decreases, since the outermost electrons are far away from the nucleus of an atom and so attains greater shielding effect. They encounter a weaker force of attraction to the +ve charge on the nucleus. And so, Ionization energy decreases from top to bottom in a group.

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Which is a way you would expect two animals of the same species to differ?

LuckyWell [14K]

C is the correct answer

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3 years ago

A sample of iron with a mass of 200g releases 9840 cal when it freezes at it’s freezing point. What is the molar heat of fusion

antoniya [11.8K]

Answer:-

2747.7 Cal mol -1

Explanation:-

Molar heat of Fusion is defined as the amount of heat necessary to melt (or freeze) 1 mole of a substance at its melting point.

Atomic mass of Iron = 55.845 g mol-1

Mass of Iron = 200 g

Number of moles of Iron = 200 g / (55.845 g mol-)

= 3.581 moles

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3 years ago

Read 2 more answers

Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.

Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

$T = 74$

The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

$E = 1 * 40$

$E = 40$

So, the percentage component of calcium is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{40}{74} * 100\%$

$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{2}{74} * 100\%$

$\%C_E = \frac{200}{74}\%$

$\%C_E = 2.70\%$

Similarly, for oxygen:

The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

$E = 32$

So, the percentage component of oxygen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{32}{74} * 100\%$

$\%C_E = \frac{3200}{74}\%$

$\%C_E = 43.24\%$

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Explanation:

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