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Serga [27]
3 years ago
9

In the three-dimensional structure of methane, ch4, the hydrogen atoms attached to a carbon atom are aligned ________. in the th

ree-dimensional structure of methane, ch4, the hydrogen atoms attached to a carbon atom are aligned ________. at the corners of a square at the corners of a tetrahedron at the corners of a cube in a straight line at the corners of a rectangle

Chemistry
1 answer:
Nikitich [7]3 years ago
4 0
Answer:
            In the three-dimensional structure of methane, CH₄, the hydrogen atoms attached to a carbon atom are aligned <span><u>at the corners of a Tetrahedron</u>.

Explanation:
                   In Methane (CH</span>₄) the central atom (carbon) is bonded to four hydrogen atoms. So, the central atom is having four electron pairs and all pairs are bonding pairs and lacks any lone pair of electron. According to Valence Shell Electron Pair Repulsion (VSEPR) Theory the central atom with four bonding pair electrons and zero lone pair electrons will attain a tetrahedral geometry with bond angles of 109°. Hence, the shape of CH₄ is tetrahedral.

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What is the identity of a 100. g sample of metal that, upon absorbing 4680 J of heat, increases in temperature by 52.0°C?
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The metal is aluminium

<u>Explanation:</u>

Given:

Heat, q = 4680 J

Mass, m = 100g = 0.1kg

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We know:

q = mcΔT

On substituting the value we get:

4680 = 0.1 X c X(52)\\\\c = \frac{4680}{5.2} \\\\c = 900 J/Kg^oC

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What is the empirical formula of a compound with a percent composition of 22.5% Phosphorous and 77.5% Chlorine?
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Answer:

\boxed {\boxed {\sf PCl_3}}

Explanation:

We are given the percent composition: 22.5% phosphorus and 77.5% chlorine.

We can assume there are 100 grams of this compound. We choose 100 because we can simply use the percentages as the masses.

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Next, convert these masses to moles, using the molar masses found on the Periodic Table.

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Use the molar masses as ratios and multiply by the number of grams. 22.5 \ g \ P  * \frac {1 \ mol \ P }{30.974 \ g \ P}= \frac {22.5 \ mol \ P }{ 30.974} = 0.7264157035 \ mol \ P

77.5 \ g \ Cl  * \frac {1 \ mol \ Cl }{35.45 \ g \ Cl}= \frac {77.5 \ mol \ Cl }{ 35.45} \ =2.186177715 \ mol \ Cl

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\frac {0.7264157035} {0.7264157035} = 1

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The mole ratio is about 1 P: 3 Cl, so the empirical formula is written as:<u> PCl₃</u>

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