Question 3 (0.1 points)

The solubility of NaCH3CO2 in water is ~1.23 g/mL. What would be the best method for preparing a supersaturated NaCH3CO2 solutio

Len [333]

Answer:

b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.

Explanation:

The solubility of NaCH₃CO₂ in water is ~1.23 g/mL. This means that at room temperature, we can dissolve 1.23 g of solute in 1 mL of water (solvent).

What would be the best method for preparing a supersaturated NaCH₃CO₂ solution?

a) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at room temperature while stirring until all the solid dissolves. NO. At room temperature, in 100 mL of H₂O can only be dissolved 123 g of solute. If we add 130 g of solute, 123 g will dissolve and the rest (7 g) will precipitate. The resulting solution will be saturated.

b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature. YES. The solubility of NaCH₃CO₂ at 80 °C is ~1.50g/mL. If we add 130 g of solute at 80 °C and let it slowly cool (and without any perturbation), the resulting solution at room temperature will be supersaturated.

c) add 1.23 g of NaCH₃CO₂ to 200 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature. NO. If we add 1.23 g of solute to 200 mL of water, the resulting solution will have a concentration of 1.23 g/200 mL = 0.00615 g/mL, which represents an unsaturated solution.

5 0

3 years ago

Show with dot and cross diagram of the formation of ions in the reaction of Fluorine with sodium ions.

Kipish [7]

Answer:

Electron dot diagram is attached below

Explanation:

Sodium is alkali metal and present in group one. It has one valence electron. All alkali metal form salt when react with halogens.

Sodium loses its one electron to get stable. While all halogens have seven valence electrons they need only one electron to get stable electronic configuration.

When alkali metals such as sodium react with halogen fluorine it loses its one valence electron which is accepted by fluorine and ionic bond is formed. The compound formed is called sodium fluoride.

Na + F → NaF

In cross and dot diagram electrons of one atom are shown as dots while other atom shown as cross to distinguish.

Electron dot diagram is attached below.

Download pdf

3 0

3 years ago

78.6 grams of O2 and 67.3 grams of F2 are placed in a container with a volume of 40.6 L. Find the total pressure if the gasses a

saul85 [17]

1) List the known and unknown quantities.

Sample: O2.

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Sample: F2.

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

2) Find the pressure of O2.

2.1- List the known and unknown quantities.

Sample: O2.

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

2.2- Convert grams of O2 to moles of O2.

The molar mass of O2 is 31.9988 g/mol.

$mol\text{ }O_2=78.6\text{ }g*\frac{1\text{ }mol\text{ }O_2}{31.9988\text{ }g\text{ }O_2}=2.46\text{ }mol\text{ }O_2$

2.3- Set the equation.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

2.4- Plug in the known quantities and solve for P.

$(P)(40.6\text{ }L)=(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)$

.

$P_{O_2}=\frac{(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)}{40.6\text{ }L}$ $P_{O_2}=1.57\text{ }atm$

The pressure of O2 is 1.57 atm.

3) Find the pressure of F2.

3.1- List the known and unknown quantities.

Sample: F2.

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

3.2- Convert grams of F2 to moles of F2.

The mmolar mass of F2 is 37.9968 g/mol.

$mol\text{ }F_2=67.3\text{ }g\text{ }F_2*\frac{1\text{ }mol\text{ }F_2}{37.9968\text{ }g\text{ }F_2}=1.77\text{ }mol\text{ }F_2$

3.3- Set the equation.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

3.4- Plug in the known quantities and solve for P.

$(P)(40.6\text{ }L)=(1.77\text{ }mol\text{ }F_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)$

.

$P_{F_2}=\frac{(1.77molF_2)(0.082057L*atm*K^{-1}*mol^{-1})(316.28K)}{40.6\text{ }L}$ $P_{F_2}=1.13\text{ }atm$

The pressure of F2 is 1.13 atm.

4) The total pressure.

Dalton's law - Partial pressure. This law states that the total pressure of a gas is equal to the sum of the individual partial pressures.

4.1- Set the equation.

$P_T=P_A+P_B$

4.2- Plug in the known quantities.

$P_T=1.57\text{ }atm+1.13\text{ }atm$ $P_T=2.7\text{ }atm$

The total pressure in the container is 2.7 atm.

5 0

1 year ago

A 30.5 gram sample of glucose (c6h12o6) contains ________ atoms of oxygen.

Margaret [11]

First, find the number of moles of glucose in 30.5 g.

You need the molar mass.

Molar mass of C6H12O6 = 6 * 12.01 g/mol + 12 * 1.01 g/mol + 6*16g/mol = 180.2 g/mol

number of moles = mass in grams / molar mass = 30.5 g / 180.2 g/mol = 0.169 mol

Second, use molecular formula

1 mol of C6H12O6 contains 6 moles of O.

Then, 0.169 moles of C6H12O6 contains 6 *0.169 = 1.014 moles of O.

Third, multiply by Avogagro's number:

1.014 moles * 6.022 * 10^23 atoms / mole = 6.106 * 10^ 23 atoms.

Answer: 6.106 * 10 ^23 atoms of oxygen.

8 0

3 years ago

Read 2 more answers

Which of the following statements regarding the importance of mineral resources is not true?

pishuonlain [190]

Answer:

They affect the weathering of molecules and atoms.

Explanation:

3 0

3 years ago