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Zielflug [23.3K]
3 years ago
13

Identify the atom H₄ CH₄ N

Chemistry
1 answer:
marysya [2.9K]3 years ago
3 0

Answer:

Organic Molecule.

Explanation:

Any molecule with C atoms and H atoms are organic molecules<em> </em>in chemistry.

Examples could be C₄H₅, C₉H₁₈O₆, C₅H₁₂N.

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Since vinegar is 5% acetic acid and 95% water, find the mole ratio of acetic acid to water in 100 g of vinegar? Hint: Instead of
Oksanka [162]

The mole ratio of acetic acid to water in 100 g of vinegar is 0.015 : 0.985.

<h3>What is the mole ratio of acetic acid to water in 100 g of vinegar?</h3>

The mole ratio of acetic acid to water in 100 g of vinegar is determined from their percentage composition.

The percentage composition of acetic acid and water in vinegar is 5% acetic acid and 95% water.

In 100 g of vinegar, there are 5 g of acetic acid and 5 g of water.

Moles = mass/molar mass

molar mass of acetic acid = 62 g/mol

molar mass of water = 18 g/mol

moles of vinegar = 5/62 = 0.08

moles of water = 95/18 = 5.28

total moles = 5.36

Mole ratio of vinegar to water = 0.08/5.36 : 5.28/5.36

Mole ratio of vinegar to water = 0.015 : 0.985

In conclusion, the mole ratio is determined from the percentage composition of acetic acid and water in vinegar.

Learn more about mole ratio at: brainly.com/question/19099163

#SPJ1

7 0
1 year ago
The parallel around the middle of earth
Vika [28.1K]
The equator is the parallel
7 0
3 years ago
What are some signs of cyanide poisoning visible on a corpse?
lina2011 [118]
Hello!
I saw this question and instantly knew I could help. I recently took a course on toxic gasses and poisons. Here's what I know.

It can be swallowed, inhaled, or absorbed through skin. It is generally released from its host compound by acids, such as the hydrochloric acid found in the stomach. The poison in the seeds is released only if the seeds are chewed.

Effects and symptoms:
Cyanide prevents the red blood cells from absorbing oxygen. It's called chemical asphyxia.
Smelling of a toxic dose of the gas can cause immediate unconsciousness, convulsions and death within one to fifteen minutes.
If swallowed a fatal dose can take up to twenty minutes or longer, esp. if swallowed on a full stomach.
If a near-lethal dose is absorbed through the skin, inhaled or swallowed the symptoms will include gasping for breath, dizziness, flushing, headache, nausea, vomiting, rapid pulse, and a drop in blood pressure causing fainting.
<span>With a lethal dose, convulsions with in four hours, except in the case of sodium nitroprusside, when death can be delayed as long as 12 hours after ingestion. </span>The victims blood may appear purple or cherry red, as in carbon monoxide poisoning, and the corpse may have pinker than normal skin.
<span>the famous bitter almond odor can be a clue and maybe noticeable at autopsy, but not everyone is capable of smelling it.

Hope this helped! :)</span>
5 0
2 years ago
Write the cell notation for an electrochemical cell consisting of an anode where Mn (s) is oxidized to Mn2 (aq) and a cathode wh
morpeh [17]

Answer:

Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

Explanation:

In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.

For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.

Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

8 0
3 years ago
The empirical formula of an organic compound is C2H4O. The molecular mass of the compound is 176g/mol.
Brrunno [24]

Answer:

The molecular formula of the compound is C_{8}H_{16}O_{4}. The molecular formula is obtained by the following expression shown below

\textrm{Molecular formula }= n\times \textrm{Empirical formula}

Explanation:

Given molecular mass of the compound is 176 g/mol

Given empirical formula is  C_{2}H_{4}O

Atomic mass of carbon, hydrogen and oxygen are 12 u , 1 u and 16 u respectively.

Empirical formula mass of the compound = \left ( 2\times12+4+16 \right ) \textrm{ u} = 44 \textrm{ g/mol}

n = \displaystyle \frac{\textrm{Molecular formula mass}}{\textrm{Empirical formula mass}} \\n = \displaystyle \frac{176}{44} = 4

\textrm{Molecular formula }= n\times \textrm{Empirical formula}

Molecular formula = 4 \times C_{2}H_{4}O

Molecular formula is C_{8}H_{16}O_{4}

6 0
2 years ago
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