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zlopas [31]
3 years ago
14

Are molecules with identical chemical formulas but different three-dimensional structures

Chemistry
1 answer:
Radda [10]3 years ago
5 0
<span>Molecules are having the same chemical formula same number of atom but different three dimensional shapes are called <u>isomer</u>. Cis trans isomers maintain the same covalent partnerships,but atom may arranged differently.There are two or more compounds arranged in same molecular formula but different arranged atoms are seen in this molecule with different structure of it.</span>
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Module 2 dba for chemistry questions? flvs
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3 years ago
A 250 cm^3 solution containing 1,46 g of sodium chloride is added to an excess of silver nitrate solution. The reaction is given
faust18 [17]

Answer:

The mass of the precipitate  that AgCl is 3.5803 g.

Explanation:

a) To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of solute (NaCl) = 1.46 g

Molar mass of sulfuric acid = 58.5 g/mol

Volume of solution = 250 cm^3 =250 mL

1 cm^3= 1 ml

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{1.46g\times 1000}{58.5g/mol\times 250}\\\\\text{Molarity of solution}=0.09982 M

0.09982 M is the concentration of the sodium chloride solution.

b) NaCl (aq)+AgNO_3 (aq)\rightarrow AgCI(s)+NaNO_3(aq)

Moles of NaCl = \frac{1.46 g}{58.5 g/mol}=0.02495 mol

according to reaction 1 mol of NaCl gives 1 mol of AgCl.

Then 0.02495 moles of NaCl will give:

\frac{1}{1}\times 0.02495 mol=0.02495 mol of AgCl

Mass of 0.02495 moles of AgCl:

0.02495 mol\times 143.5 g/mol=3.5803 g

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3 years ago
What is the mass of 0.5 moles of carbon tetrafluoride, CF4?
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Answer:

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The formula for the number of moles (n) is equal to n=\frac{mass}{molecular weight} .

Since we need to find the mass, we derive it from the formula of the number of moles and we get that mass = n x molecular weight .

The molecular weight of CF_{4} = 12 g/mol (from the carbon) + 19x4 g/mol (from the 4 fluorine atoms)= 88 g/mol

We plug in the numbers in the derived formula for the mass and we get :

mass = n x molecular weight = 0.5 mol x 88 g/mol = 44 g

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