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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.

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Answer:

the first one

Explanation:

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The order increasing acidity of the solutions is; CH3COONa ≈ NaF < CH3COONH4 < NH4NO3.

An acidic solution is one whose pH lies between 0 to 6. A salt that undergoes hydrolysis to give an acid will have an acidic pH. Let us now consider the solutions to know which of them will give an acid solution.

• The salt NH4NO3 will give an acid solution because its hydrolysis yields nitric acid
• The salt  CH3COONH4 yields a neutral solution
• The salt NaF yields a basic solution because its hydrolysis yields NaOH.
• The salt CH3COONa yields a basic solution since its hydrolysis yields NaOH.

The order increasing acidity of the solutions is; CH3COONa ≈ NaF < CH3COONH4 < NH4NO3.

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Answer:

The answer to the question is

In order of increasing acidity, we have

CH3COONa→NaF→NaNO3→CH3COONH4→NH4NO3

Explanation:

(i) NH4NO3 is an acidic salt as it consists of NH₃ and HNO₃. NH₃ is a weak base and  HNO₃ is a strong acid pH = 5.4

(ii) NaNO3  PH7 produces a neutral solution when dissolved in water

(iii) CH3COONH4 Very concentrated solutions are acidic while low concentration 0.5 M are neutral, pH = 7

(iv) NaF basic as it is formed from  weak acid and  strong base. The strength of the basicity depends on the solution pH 7.4

(v) CH3COONa Basic in small concentration pH  8 or 9

Therefore we have in order of increasing acidity, CH3COONa→NaF→NaNO3→CH3COONH4→NH4NO3