Two different atoms have six protons each and the same mass. However, one has a negative charge while the other has a positive c

1) The chemical formula for propane is CH₃-CH₂-CH₃. 
Propane is classified as a three-carbon alkane (acyclic saturated <span>hydrocarbon).
</span>2) The chemical formula for propanal is CH₃-CH₂-CH=O.
Propanal <span> is a </span>saturated<span> three-carbon </span>aldehyde (consists of<span> a </span>carbonyl<span> center).
3) </span>The chemical formula for propanol is CH₃-CH₂-CH₂-OH.
1-propanol <span> is a </span><span>primary alcohol.
4) </span>The chemical formula for propanone is (CH₃)₂-C=O.
Propanone, also known as acetone, is <span>the simplest and smallest</span> ketone.

Answer:

The molality of the solution is 1.08 m.

Explanation:

First, determine the mass of the solvent.

A 13% solution by mass indicates that 13 grams are found in every 100 grams of solution.

Thus, solution mass = solute mass + solvent mass

100 g = 13 g + solvent mass

Therefore, solvent mass = 100 g - 13 g → 87 g

Next, we calculate the moles of solute (mass / molar mass):

13 g / 138.2 g/mol = 0.094 moles

Finally, to find the molality, which is the moles of solute per 1 kg of solvent (mol/kg), we convert the solvent mass to kg:

87 g. 1 kg / 1000 g = 0.087 kg

Then, molality → 0.094 mol / 0.087 kg = 1.08 m

Answer:

The solution composed of 35.0 g of mixed with 250.0 g of ethanol will exhibit the lowest freezing temperature

Explanation:

where,

= reduction in freezing point =  

= freezing point constant  

m = molality

As observed, a higher molality in the solution corresponds to a greater depression in the freezing point, resulting in a more reduced freezing point of the solution.

A. 35.0 g of in 250.0 g of ethanol.

The amount of moles of =

The mass of ethanol solvent is 250.0 g = 0.25 kg (1 g = 0.001 kg)

B. 35.0 g of in 250.0 g of ethanol

Moles of

The mass of ethanol solvent is 250.0 g = 0.25 kg (1 g = 0.001 kg)

C. 35.0 g of in 250.0 g of ethanol

Moles of =

The mass of ethanol solvent is 250.0 g = 0.25 kg (1 g = 0.001 kg)

The solution containing 35.0 g of added to 250.0 g of ethanol will yield the lowest freezing point

The amount of pure present in 7.5 g of equals 3.0 g.

195 g×

1 mol

Ni(OH)

2

92.71

g Ni(OH)

2

×

6.022×

10

23

formula units

mol

=1.27×

10

24

formula units of Ni(OH)

2

Given that there are  

5

atoms—one atom of  

Ni

, two of  

O

and two of  

H

—in every formula unit of  

Ni(OH)

2

, we now multiply the previously calculated formula units by  

5

:

1.27×

10

24

formula units of Ni(OH)

2

×

5 atoms

per formula unit of Ni(OH)

2

=6.33× 10^ 24  total atoms