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kati45 [8]
3 years ago
13

Consider two compounds. Compound A contains 15.7 g of sulfur and 18.6 g of fluorine. Compound B contains 25.4 g of sulfur and 60

.2 g of fluorine. For Compound A, the ratio of fluorine to sulfur is 1.18. For Compound B, the ratio of fluorine to sulfur is 2.37. Using the Law of Definite Proportions (also called the Law of Constant Composition), could Compound A and Compound B be the same compound
Chemistry
1 answer:
Lunna [17]3 years ago
4 0

Answer:

No, compound A and B are not the same compound

Explanation:

According to the law of definite proportion "every chemical compound contains fixed and constant proportions (by mass) of its constituent elements." (Encyclopedia Britannica)

We can see in the question that the ratio of flourine to sulphur in compound A is 1.18 while the ratio of flourine to sulphur in compound B is 2.37.

The two chemical compounds do not contain a fixed proportion by mass of their constituent elements therefore, they can not be same compound according to the law of definite proportions.

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Calcium chloride is salt that completely dissociate in water on calcium cations and chlorine anions.
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among these properties of acids basses or both which property is specific to acids only? A. accepts protons B. bitter taste C. e
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1 year 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

The mass of the precipitate  that AgCl is 3.5803 g.

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3 years ago
A certain FM radio wave has a frequency of 1.31 x 108 Hz. Given that radio waves travel at
Westkost [7]

Answer:

2.28 m

Explanation:

Use the relationship  

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λ=vf

 

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