<u>Answer:</u> The empirical formula for the given compound is 
<u>Explanation:</u>
We are given:
Percentage of H = 5.80 %
Percentage of O = 23.02 %
Percentage of N = 20.16 %
Percentage of Cl = 51.02 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of H = 5.80 g
Mass of O = 23.02 g
Mass of N = 20.16 g
Mass of Cl = 51.02 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Hydrogen = 
Moles of Oxygen = 
Moles of Nitrogen = 
Moles of Chlorine = 
- <u>Step 2:</u> 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 1.44 moles.
For Hydrogen = 
For Oxygen = 
For Nitrogen =
For Chlorine =
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of H : O : N : Cl = 4 : 1 : 1 : 1
Hence, the empirical formula for the given compound is
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Answer:
Explanation:
C decays by a process called beta decay. During this process, an atom of 14C decays into an atom of 14N, during which one of the neutrons in the carbon atom becomes a proton. This increases the number of protons in the atom by one, creating a nitrogen atom rather than a carbon atom.
Answer:
4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium
Explanation:
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:
- HCl: 2 moles
- Na: 1 mole
- NaCl: 2 moles
- H₂: 1 mole
You know the following masses of each element:
- H: 1 g/mole
- Cl: 35.45 g/mole
- Na: 23 g/mole
So, the molar mass of each compound participating in the reaction is:
- HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
- Na: 23 g/mole
- NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
- H₂: 2* 1 g/mole= 2 g/mole
Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:
- HCl: 2 moles* 36.45 g/mole= 72.9 g
- Na: 1 mole* 23 g/mole= 23 g
- NaCl: 2 moles* 58.45 g/mole= 116.9 g
- H₂: 1 mole* 2 g/mole= 2 g
So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?
mass of H₂= 4.8 g
<u><em>4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium</em></u>
