Explain, in detail, how you convert grams of one substance to grams of something else. Be specific and include each step

1 answer:

5 0

I will present a simple reaction so we can do this conversion:

2H₂ + O₂ → 2H₂O

We will assume we have 32 g of O₂ and we want to find the amount of water, assuming this reaction goes to completion. We must first convert the initial mass to moles, which we do using the molar mass in units of g/mol. The molar mass of O₂ is 32 g/mol.

32 g O₂ ÷ 32 g/mol = 1 mole O₂.

Now that we have moles of oxygen, we use the molar coefficients to find the ratio of water molecules to oxygen molecules. We can see there are 2 moles of water for every 1 mole of oxygen.

1 moles O₂ x (2 mol H₂O/ 1 mol O₂) = 2 moles H₂O

Now that we have the moles of water, we can convert this amount into grams using the molar mass of water, which is 18 g/mol.

2 moles H₂O x 18 g/mol = 36 g H₂O

Now we have successfully converted the mass of one molecule to the mass of another.

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Methylene chloride (CH2Cl2) has fewer chlorine atoms than chloroform (CHCl3). Nevertheless, methylene chloride has a larger mole

Talja [164]

Answer:

Explanation has been given below.

Explanation:

Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.