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Scilla [17]
3 years ago
12

The number of unique open-chain structures corresponding to the molecular formula c3h5cl is:

Chemistry
1 answer:
rosijanka [135]3 years ago
6 0

The number of unique open-chain structures corresponding to the molecular formula C₃H₅Cl is two.

<em>Step 1</em>. Draw <em>all possible connections of three carbon atoms</em>.

The is only one possibility (See first image below).

<em>Step 2</em>. Put the <em>Cl in all possible unique positions</em>.

There are only <em>two</em> possibilities: on an end carbon or on the middle carbon (see second image below).

Thus, there are only two isomers of C₃H₅Cl.

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1.00 M CaCl2 Density = 1.07 g/mL
Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

m = 1 mol\times 111 g/mol = 111 g

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

M=d\times V=1.07 g/mL\times 1000 mL=1,070 g

1) The value of %(m/M):

\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%

2) The value of %(m/V):

\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%

Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}

Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}

n = Equivalent mass

n = \frac{\text{molar mass of ion}}{\text{charge on an ion}}

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 CaCl_2 mole has 1 mole of calcium ion)

n=\frac{40 g/mol}{2}=20

=\frac{1 mol}{20 g/mol\times 1L}=0.050 N

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 CaCl_2 mole has 2 mole of chlorine ion)

n=\frac{35.5 g/mol}{1}=35.5

=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N

Moles of calcium chloride = 1.00 mol

Mass of solvent =  Mass of solution - mass of solute

= 1,070 g - 111 g = 959  g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

\frac{1 mol}{0.959 kg}=1.043 mol/kg

Moles of calcium chloride = n_1=1mol

Mass of solvent = 959 g

Moles of water = n_2=\frac{959 g}{18 g/mol}=53.28 mol

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842

7) Mole fraction of water =

\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

m'=d\times v=1.07 g/ml\times 100 g= 107 g

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

11.1\%=\frac{m}{100 mL}\times 100

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

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3 years ago
For the procedural error, indicate if the error will affect the actual yield of copper(II) saccharinate product and if it does,
jolli1 [7]

Answer:

Lowers the actual yield

Explanation:

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Angiosperms produce brightly colored blooms and sweet-smelling flowers. Why have angiosperms developed these adaptations? A. to
blsea [12.9K]

Answer:

C

Explanation:

Angiosperms have developed these adaptations because it attracts pollinators which helps the ecosystem grow.

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10. Which would not be a good solution to counter overpopulation?
Vikki [24]

Answer:

A. Encourage an increase in pollution in these countries

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You have a beaker of water that potentially has Ag+ ions and Ba+ ions (it could have both, one of them, or none of them). How wo
xenn [34]

Answer:

See explanation

Explanation:

Qualitative analysis in chemistry is a method used to determine the ions present in a solution chiefly by means of chemical reactions.

In this case, I suspect the presence of silver ions and/or barium ions. The first step is to add dilute HCl. This will lead to the precipitation of the silver ion as AgCl. If a white precipitate is formed upon addition of HCl then Ag^+ is present in the solution.

Secondly, I add a carbonate such as NH4CO3(aq). This will cause the barium ions to become precipitated as barium carbonate. Hence, the formation of a white precipitate when NH4CO3(aq) is added to the solution indicates the presence of barium ion in the solution.

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