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mezya [45]
3 years ago
12

A material being examined is the same throughout. The material contains carbon and oxygen, chemically combined. What type of mat

erial is it?
Chemistry
1 answer:
a_sh-v [17]3 years ago
3 0

Answer:

compound

Explanation:

A compound is a substance that consists of atoms of two or more various elements such that the atoms are chemically joined together.

For example, water is a compound made up of hydrogen and oxygen.

A material being examined is the same throughout such that the material contains carbon and oxygen that are chemically combined. This material is compound.

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The northern slope will have the least precipitation if winds blow against a mountain from the south.

<h3>What is Slope?</h3>

This is referred to as the degree of steepness on a surface such as rocky areas etc.

When wind blows on the southern part of a mountains, it gets stuck which is why the northern slope have the least precipitation.

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Ionic bonds form between which two types of elements?​
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Answer:

Ionic bonds usually occur between metal and nonmetal ions. For example, sodium (Na), a metal, and chloride (Cl), a nonmetal, form an ionic bond to make NaCl.

Explanation:

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

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