A saturated solution of NaNO3 is prepared at 60.°C using 100. grams of water. As this solution is cooled to 10.°C, NaNO3 precipitates (settles)out of the solution. The resulting solution is saturated. Approximately how many grams of NaNO3 settled out of the original solution?
(1) 46 g (3) 85 g
(2) 61 g (4) 126 g

Answer is: D. substitution.

Substitution reaction or single displacement reaction is a chemical reaction in which one functional group in a chemical compound is replaced by another functional group.

Balanced chemical reaction: CH₃Br(aq) + OH⁻(aq) → CH₃OH(aq) + Br⁻(aq).

In this balanced chemical reaction hydroxy group OH⁻ replaced one atom of bromine (Br⁻) in methyl bromide (CH₃Br).

Answer : The correct option is, (D) Substitution reaction.

Explanation :

Addition reaction : It is a type of reaction in which a molecule combined with the another molecule to give a new larger molecule.

The general representation of this reaction is,

Condensation reaction : It is a type of reaction in which two molecules combine to form a larger molecule and also producing a smaller molecule such as water.

Elimination reaction : It is a type of reaction in which the loss of elements  from the starting material to form a double bond in the product.

The general representation of this reaction is,

Substitution reaction : It is a type of reaction in which one functional group is replaced by the another functional group.

The general representation of this reaction is,

(B and C are the two different functional groups)

In the given reaction, Br (functional group) is replaced by the hydroxide ion (functional group). So, this reaction is a substitution reaction.

Hence, the given reaction is a substitution reaction.

Answer: A solid-state component with three layers of semiconductors is a transistor.

Explanation:

A device with the help of which current or voltage is regulated and for electronic signals it acts as a switch or gate is known as a transistor.

A transistor has three layers of semiconductors. Each of these layers are capable of carrying current.

Thus, we can conclude that a solid-state component with three layers of semiconductors is a transistor.

The enthalpy change (ΔH) for the neutralization of 0.1 moles of 1.0 M NaOH with 0.1 moles of 1.0 M HCl in a coffee-cup calorimeter is approximately 28.05 kJ/mol.

To calculate the enthalpy change (ΔH) for the neutralization of HCl by NaOH, you can use the equation:

ΔH = q / moles of limiting reactant

First, let's find the moles of the reactants. We have 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl. Since we know the volumes and concentrations, you can find the moles of each reactant using the formula:

moles = (volume in L) × (concentration in mol/L)

For NaOH:

moles of NaOH = (100.0 mL / 1000 mL/L) × 1.0 mol/L = 0.1 moles

For HCl:

moles of HCl = (100.0 mL / 1000 mL/L) × 1.0 mol/L = 0.1 moles

Now, you need to determine the limiting reactant. The balanced chemical equation for the neutralization of HCl by NaOH is:

NaOH + HCl → NaCl + H₂O

The stoichiometric ratio of NaOH to HCl is 1:1, which means they react in a 1:1 ratio. Since both reactants have 0.1 moles, neither is in excess. Therefore, the reactant that limits the reaction is the one that is present in the smaller amount, which is NaOH in this case.

Now, calculate the heat absorbed or released (q) using the equation:

q = mΔTC

Where:

m is the mass (in grams) of the solution, which we can calculate using the density of 1.0 g/cm³ and the volume (in mL).

ΔT is the change in temperature.

C is the specific heat capacity (given as 4.18 J/g°C).

For the volume of 100.0 mL, the mass is 100.0 g (since 100.0 mL = 100.0 g, given the density is 1.0 g/cm³).

ΔT = Final temperature - Initial temperature

ΔT = 31.38°C - 24.68°C = 6.70°C

Now, calculate q for the reaction:

q = 100.0 g × 6.70°C × 4.18 J/g°C = 2804.76 J

Finally, calculate the enthalpy change (ΔH) by dividing q by the moles of the limiting reactant:

ΔH = 2804.76 J / 0.1 moles = 28047.6 J/mol

Since the enthalpy change is typically expressed in kJ/mol, divide by 1000 to convert J to kJ:

ΔH = 28.05 kJ/mol

So, the enthalpy change for the neutralization of HCl by NaOH is approximately 28.05 kJ/mol.

To know more about moles:

brainly.com/question/34302357

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