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3 years ago

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Compounds may be represented by chemical formulas. In the chemical formula for water which is H20, the 2 which represents the nu

mber of hydrogen atoms is called_______?
A) the reactant.
B) the subscript.
C) the coefficient.
D) the atom factor.

2 answers:

sergey [27]3 years ago

7 0

The answer is B.....

Fed [463]3 years ago

3 0

The answer is choice "B", the subscript. Hope this helped :)

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What do you call A compound composed of two oppositely charged elements?

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Answer: Ionic Bond

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When can a compound be treated like an element when balancing chemical equations?

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A compound can be treated like an element when balancing chemical equations when : A. when there is no change to the atoms in the compound

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If the same amount of heat is suppliedto all samples of 10.0g each of aluminum iton and copper all at 15 degree celsius which sa

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When a substance absorbs thermal energy, it partitions some as potential and some as kinetic energy. Specific heat is an expression related to the quantity of heat a substance stores as potential energy; the remainder is absorbed as kinetic which causes the temperature to increase - recall that temperature is a measure of average kinetic energy.
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3 years ago

A 17.11 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 21.71 g CO2 and 5.

Andru [333]

Answer: The empirical formula and the molecular formula of the organic compound is $CHO$ and $C_4H_4O_4$ respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ = 21.71 g

Mass of $H_2O$ = 5.926 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.71 g of carbon dioxide, = $\frac{12}{44}\times 21.71=5.921g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.926 g of water, = $\frac{2}{18}\times 5.926=0.658g$ of hydrogen will be contained.

Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53 g

Mass of C = 5.921 g

Mass of H = 0.658 g

Mass of O = 10.53 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.921g}{12g/mole}=0.493moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.658g}{1g/mole}=0.658moles$

Mass of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{10.53g}{16g/mole}=0.658moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.493}{0.493}=1$

For H = $\frac{0.658}{0.493}=1$

For O= $\frac{0.658}{0.493}=1$

The ratio of C : H: O = 1: 1: 1

Hence the empirical formula is $CHO$ .

empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29

Molecular mass = 104.1 g/mol

$n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=\frac{104.1}{29}=4$

Thus molecular formula = $n\times {\text {Empirical formula}}=4\times CHO=C_4H_4O_4$

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Why is it important to use standardized safety symbols on all hazardous products

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It's simply, e.g. I'm from Poland, imagine that you give me strong acid (HF, dangerous corrosive substance), so how can I recognize that this substance is dangerous? Safety symbols on all hazardous products are <span>requirement in chemistry.</span>

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