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Answer:phone

Explanation:phone

Answer:

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Answer:

Weather forecasting

Explanation:

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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Answer:

B

Explanation:

A) The sea of electrons only happen with metalic bond, a bond with two metalic elements.

B) As both elements in ammonia (N and H) are non-metals we don't observe an transfer of electrons, both elements share the number of electrons needed to stablize it (3 electrons for the N and 1 for the H).

C) To have a transfer of electrons, we need a bond between a metal and a non-metal, in this case, N and H are non-metals, so this doesn't happen.

D) The electrons are shared, only the electrons nothing happens between the orbitals.

How is electron movement related to covalent bonding in ammonia, NH3?

A) Electrostatic sharing in a sea of electrons around the atoms allows bonds to form. 
B) The atomic orbitals overlap and electrons are shared between the atoms forming bonds. 
C) A transfer of electrons forms ions which are electrostatically attracted forming bonds. 
D) Orbital exchange occurs between the atoms to redistribute the electrons and form bonds.

The electron movement related to covalent bonding in ammonia, NH3 is The atomic orbitals overlap and electrons are shared between the atoms forming bonds. The answer is letter B.