An article about half-lives describes a daughter isotope. What is a daughter isotope?

Answer:

Molecular bonding cannot occur without electrons.

Explanation:

In order to be 100% correct, the answers should be electrons, protons and neutrons. Since moleculas are formed by atoms and atoms are formed by electrons protons and neutrons.

However, if we consider only the bonding itself, the most important atomic particle is the electron. Molecular or convalent bonding happens when atoms share electrons pairs to reach stability.

Answer:

-4741 °C

Something is strange, because this is a weird number.

Explanation:

ΔT = Kf . m. i

That's the colligative property of freezing point depression.

Kf = Cyroscopic constant

m = molality (moles of solute in 1kg of solvent)

i = Van't Hoff factor (numbers of ions dissolved)

We assume 100% dissociation:

CaCl₂ → Ca²⁺ + 2Cl⁻   i = 3

ΔT = Freezing point of pure solvent - Freezing point of solution

Let's determine molality

Solute = CaCl₂

Moles of solute = 23.5 g . 1 mol/ 110.98 g = 0.212 moles

We determine the mass of solvent by density. Density's data is in g/L. We need to convert the volume from mL to L

250 mL . 1L / 1000 mL = 0.250 L

0.997 g/L = mass of water / volume of water → 0.997 g/L . volume of water = mass of water

0.997 g/L . 0.250L = 0.249 g

Now, we convert the mass of water from g to kg

0.249 g . 1 kg / 1000 g = 2.49×10⁻⁴ kg

Molality = mol/kg → 0.212 mol / 2.49×10⁻⁴ kg = 850.5 m

We replace data:

0°C - Freezing point of solution = 1.858 °C . kg /mol . 850.5 mol/kg . 3

Freezing point of solution = -4741 °C

Answer:

The freezing point of the solution is -4.74 °C

Explanation:

Step 1: data given

Mass of calcium chloride CaCl2 = 23.50 grams

Step 2: Calculate moles CaCl2

Moles CaCl2 = mass CaCl2 / molar mass CaCl2

Moles CaCl2 = 23.50 grams / 110.98 g/mol

Moles CaCl2 = 0.212 moles

Step 3: Calculate mass H2O

Density = mass / volume

Mass = density * volume

Mass H2O = 997 g/L * 0.250 L

MAss H2O = 249.25 grams

Step 3: calculate molality of the solution

Molality = moles CaCl2 / mass H2O

Molality = 0.212 moles / 0.24925 kg

Molality = 0.851 molal

ΔT = i*Kf*m  

⇒with ΔT = the freezing point depression = TO BE DETERMNED

⇒ with i = the van't Hoff factor = 3

⇒with Kf = the freezing point depression constant of water = 1.858 °C /m

⇒with m = the molality = moles CaCl2 / mass water = 0.851 molal

ΔT = 3 * 1.858 * 0.851

ΔT =  4.74 °C

Step 4: Calculate the freezing point of the solution

ΔT = T (pure solvent) − T (solution)

ΔT  = 0°C - 4.74 °C

The freezing point of the solution is -4.74 °C

NOTE: the density of water = 0.997 kg/L or 997 g/L

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