What is the name of the binary molecule H2O2

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Kamila [148]

Answer:

gas

Explanation:

8 0

3 years ago

Is it necessary that compounds be colored to be separated by chromatography?

blondinia [14]

No itt is not required

4 0

3 years ago

A 0.1014 g sample of a purified compound containing C, H, and, O was burned in a combustion apparatus and produced 0.1486 g CO2

Alina [70]

Answer: The empirical formula for the given compound is $CH_2O$

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 $CO_2=0.1486g$

Mass of $H_2O=0.0609g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 0.1486 g of carbon dioxide, $\frac{12}{44}\times 0.1486=0.0405g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.0609 g of water, $\frac{2}{18}\times 0.0609=0.00677$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1014) - (0.0405 + 0.00677) = 0.054 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.0405g}{12g/mole}=0.003375moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.00677g}{1g/mole}=0.00677moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.054g}{16g/mole}=0.003375moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.

For Carbon = $\frac{0.003375}{0.003375}=1$

For Hydrogen = $\frac{0.00677}{0.003375}=2.00\times 2$

For Oxygen = $\frac{0.003375}{0.003375}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 2 : 1

The empirical formula for the given compound is $C_1H_2O_1=CH_2O$

Thus, the empirical formula for the given compound is $CH_2O$

8 0

3 years ago

35.10 g of aluminum hydroxide is allowed to react with 53.94 g of sulfuric acid as follows:2Al(OH)3(s) + 3H2SO4(aq) -------->

ikadub [295]

Answer:

62.586 gram

Explanation:

moles of Al(OH)3 = mass / molar mass = 35.1 / (27+17x3) = 0.45 mol

moles of H2SO4 = mass / molar mass = 53.94 / (2+32+16x4) = 0.55 mol

H2SO4 is the limiting reagent (reacts completely)

⇒ moles of Al2(SO4)3 is worked out by moles of H2SO4

moles of Al2(SO4)3 = moles of H2SO4 / 3 = 0.183 mol

mass of Al2(SO4)3 = mole x molar mass = 0.183 x (27x2 + 96x3) = 62.586 gram

8 0

3 years ago

The last intermediate in citric acid cycle is:_________

GenaCL600 [577]

The last intermediate in citric acid cycle is Oxaloacetic acid.

<h3>What is Citric Acid Cycle?</h3>

Organic molecule HOC(CO2H)(CH2CO2H)2 is the chemical formula for citric acid. It is a weak organic acid that is colorless. Citrus fruits naturally contain it. It is a biochemical intermediary in the citric acid cycle, which is a component of all aerobic organisms' metabolism.

Every year, more than two million tons of citric acid are produced. It is frequently used as a flavoring, an acidifier, and a chelating agent.

Citrates, which include salts, esters, and the polyatomic anion present in solution, are derivatives of citric acid. Trisodium citrate is an example of the former; triethyl citrate is an example of an ester.

Learn more about citric acid with the help of the given link:

brainly.com/question/15582668

#SPJ4

7 0

1 year ago