All categories

3 years ago

Both elements and compounds are considered to be

Chemistry

1 answer:

frez [133]3 years ago

7 0

Answer:

Pure Substances.

Explanation:

Elements and compounds are both examples of pure substances. A substance that cannot be broken down into chemically simpler components is an element.

You might be interested in

The elements X and Y combine in different ratios to form four different types of compounds: XY, XY2, XY3, and XY4. Consider that

Law Incorporation [45]

Answer:

Ratios in order of increasing value ; The ratio of the mass ratio of Y to X in XY2 to the mass ratio of Y to X in XY, The ratio of the mass ratio of Y to X in XY3 to the mass ratio of Y to X in XY, The ratio of the mass ratio of Y to X in XY4 to the mass ratio of Y to X in XY

1) Mass ratio = 3

2) Mass ratio = 2

3) Mass ratio = 4

Explanation:

The detailed and step by step calculation is shown in the attachment.

3 0

3 years ago

If a compound begins with a metal, it most likely is a _______ compound

Salsk061 [2.6K]

Answer:

Explanation:

6 0

2 years ago

Methanol can be produced by the following reaction: CO(g) 2 H2(g) CH3OH(g). How is the rate of disappearance of hydrogen gas rel

belka [17]

Answer:

$r_{H_2}=-2r_{CH_3OH}$

Explanation:

Hello!

In this case, for the reaction:

$CO(g)+ 2 H_2(g) \rightarrow CH_3OH(g)$

In such a way, via the rate proportions, that is written considering the stoichiometric coefficients, we obtain:

$\frac{1}{-1} r_{CO}=\frac{1}{-2} r_{H_2}=\frac{1}{1} r_{CH_3OH}$

Whereas the reactants, CO and H2 have negative stoichiometric coefficients; therefore the rate of disappearance of hydrogen gas is related to the rate of appearance of methanol as shown below:

$\frac{1}{-2} r_{H_2}=\frac{1}{1} r_{CH_3OH}\\\\r_{H_2}=\frac{-2}{1} r_{CH_3OH}\\\\r_{H_2}=-2r_{CH_3OH}$

Which means that the rate of disappearance of hydrogen gas is negative and the rate of appearance of methanol is positive.

Regards!

7 0

3 years ago

What exactly is a lode

Aloiza [94]

Answer:

A veinlike deposit, usually metalliferous.

Any body of ore set off from adjacent rock formations.

A rich supply or source

4 0

2 years ago

Read 2 more answers

Find the number of moles of butane (c4h10) in 21g sample of butane gas

iogann1982 [59]

Answer:

0.362 moles

Explanation:

Mass of butane = 21g

Molar mass of carbon = 12g / mol

Molar mass of hydrogen = 1g/mol

Molar mass of butane ? = [(12*4) * (1*10)]

Molar mass of butane = 58g / mole

Number of moles = mass of molecules / molar mass of molecule

Number of moles = 21 / 58

Number of moles of butane = 0.362 moles

The number of moles in 21g of butane is 0.362 moles

8 0

3 years ago

Both elements and compounds are considered to be￼￼

Both elements and compounds are considered to be