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3 years ago

The chemical formula for table sugar is C12H11O22. What can you tell from this formula?

Chemistry

2 answers:

just olya [345]3 years ago

7 0

I say the answer is The ratio of oxygen atoms to hydrogen atoms in a molecule of sugar is 2 to 1

Archy [21]3 years ago

4 0

Answer: The correct answer is the ratio of oxygen atoms to hydrogen atoms in a molecule of sugar is 2 to 1.

Explanation:

The given chemical formula for table sugar is $C_{12}H_{11}O_{22}$ .

The above compound contains 12 atoms of carbon atom, 11 atoms of hydrogen and 22 atoms of oxygen.

The ratio of oxygen atoms to hydrogen atoms is $22:11::2:1$ and the ratio of carbon atoms to hydrogen atoms in a molecule of sugar is $12:11$

An oxygen atom contains 8 protons and a carbon atoms has 6 electrons.

Hence, the correct statement is the ratio of oxygen atoms to hydrogen atoms in a molecule of sugar is 2 to 1.

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The acid HOCl (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles

sergejj [24]

Answer: The expression for equilibrium constant is $K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}$

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

For the general chemical equation:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_c$ is given as:

$K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

For the given chemical reaction:

$2HgO(s)+H_2O(l)+2Cl_2(g)\rightleftharpoons 2HOCl(aq.)+HgO.HgCl_2(s)$

The expression for $K_{eq}$ is given as:

$K_{eq}=\frac{[HOCl]^2[HgO.HgCl_2]}{[HgO]^2[H_2O][Cl_2]^2}$

The concentration of solid is taken to be 0.

So, the expression for $K_{eq}$ is given as:

$K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}$

3 0

3 years ago

The process of an atom releasing energy when it moves to a lower energy state is called

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The process of an atom releasing energy when it moves to a lower energy state is called emission.

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3 years ago

What volume of water is required to prepare 0.1 M H3PO4 from 100 ml of 0.5 M solution?

ExtremeBDS [4]

Answer: A volume of 500 mL water is required to prepare 0.1 M $H_{3}PO_{4}$ from 100 ml of 0.5 M solution.

Explanation:

Given: $M_{1}$ = 0.1 M, $V_{1}$ = ?

$M_{2}$ = 0.5 M, $V_{2}$ = 100 mL

Formula used to calculate the volume of water is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.1 M \times V_{1} = 0.5 M \times 100 mL\\V_{1} = 500 mL$

Thus, we can conclude that a volume of 500 mL water is required to prepare 0.1 M $H_{3}PO_{4}$ from 100 ml of 0.5 M solution.

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3 years ago