All categories

3 years ago

Chlorine is much more apt to exist as an anion than is sodium. this is because __________.

1 answer:

s2008m [1.1K]3 years ago

6 0

The answer is that “chlorine has a greater electron affinity than sodium does”. Sodium has a lesser electron affinity than Chlorine, which is why it is deliberated that Chlorine is much more suitable to be existent as an anion than sodium, due to the lack of its electron affinity.

You might be interested in

Camphor, a white solid with a pleasant odor, is extracted from the roots, branches, and trunk of the camphor tree. Assume you di

katrin [286]

Answer: The molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

Explanation:

Calculating the molarity of solution:

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

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

Given mass of camphor = 70.0 g

Molar mass of camphor = 152.2 g/mol

Volume of solution = 575 mL

Putting values in above equation, we get:

$\text{Molarity of camphor}=\frac{70\times 1000}{152.2\times 575}\\\\\text{Molarity of camphor}=0.799M$

Calculating the molarity of solution:

To calculate the mass of ethanol, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanol = 0.785 g/mL

Volume of ethanol = 575 mL

Putting values in above equation, we get:

$0.785g/mL=\frac{\text{Mass of ethanol}}{575mL}\\\\\text{Mass of ethanol}=(0.785g/mL\times 575mL)=451.38g$

To calculate the molality of solution, we use the equation:

$\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$m_{solute}$ = Given mass of solute (camphor) = 70 g

$M_{solute}$ = Molar mass of solute (camphor) = 152.2 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 451.38 g

Putting values in above equation, we get:

$\text{Molality of camphor}=\frac{70\times 1000}{152.2\times 451.38}\\\\\text{Molality of camphor}=1.02m$

Calculating the mole fraction of camphor:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For camphor:

Given mass of camphor = 70 g

Molar mass of camphor = 152.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of camphor}=\frac{70g}{152.2g/mol}=0.459mol$

For ethanol:

Given mass of ethanol = 451.38 g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of ethanol}=\frac{451.38g}{46g/mol}=9.813mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

Moles of camphor = 0.459 moles

Total moles = [0.459 + 9.813] = 10.272 moles

Putting values in above equation, we get:

$\chi_{(camphor)}=\frac{0.459}{10.272}=0.045$ \

Calculating the mass percent of camphor:

To calculate the mass percentage of camphor in solution, we use the equation:

$\text{Mass percent of camphor}=\frac{\text{Mass of camphor}}{\text{Mass of solution}}\times 100$

Mass of camphor = 70 g

Mass of solution = [70 + 451.38] = 521.38 g

Putting values in above equation, we get:

$\text{Mass percent of camphor}=\frac{70g}{521.38g}\times 100=13.43\%$

Hence, the molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

3 0

3 years ago

"a basic experiment involves a minimum of ________ participant group(s)."

3241004551 [841]

Answer: TWO.

At least two groups: treatment group and control group.

The treatment group is that is exposed to the different levels of the independent variable ( a medication for example), while the control group is not treated, so the researchers can compare the effect of the medication.

5 0

3 years ago

The vaporization of 1 mole of liquid water (system) at 100.9 C, 1.00 atm, is endothermic.

yan [13]

Answer:

a) The work done is 10.0777 kJ

b) The water's change in internal energy is -122.1973 kJ

Explanation:

Given data:

1 mol of liquid water

T₁ = temperature = 100.9°C

P = pressure = 1 atm

Endothermic reaction

T₂ = temperature = 100°C

1 mol of water vapor

VL = volume of liquid water = 18.8 mL = 0.0188 L

VG = volume of water vapor = 30.62 L

3.25 moles of liquid water vaporizes

Q = heat added to the system = -40.7 kJ

Questions: a) Calculate the work done on or by the system, W = ?

b) Calculate the water's change in internal energy, ΔU = ?

Heat for 3.25 moles:

$Q_{1} =3.25*(-40.7)=-132.275kJ$

The work done:

$W=-nP*delta(V)=-3.25*101.33*(0.0188-30.62)=10077.6637J=10.0777kJ$

The change in internal energy:

$delt(U)=W+Q_{1} =10.0777-132.275=-122.1973kJ$

7 0

3 years ago

In a chemical equation, the arrow

muminat

In a chemical equation, the arrow

A. can be read as "yields" or "makes."

B. always points toward the products.

C. separates the products and reactants.

D. all of these

all of these options are right.

7 0

3 years ago

Read 2 more answers

What would happen to earth's orbit around the sun if there were gravity

kipiarov [429]

If the sun was not there the earth would travel in a straight line

7 0

4 years ago