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:
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Answer:2.50 moles of magnesium will consume approximately 182.30 grams of hydrochloric acid in the given reaction.
Explanation:To find out how many grams of hydrochloric acid (HCl) are consumed when 2.50 moles of magnesium (Mg) react with it, you can use stoichiometry and the balanced chemical equation:
Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)
From the balanced equation, you can see that 1 mole of magnesium (Mg) reacts with 2 moles of hydrochloric acid (HCl).
Now, let's use this information to calculate the moles of HCl required to react with 2.50 moles of Mg:
Moles of HCl = (2.50 moles Mg) * (2 moles HCl / 1 mole Mg)
Moles of HCl = 2.50 moles * 2
Moles of HCl = 5.00 moles
Now that we know we need 5.00 moles of HCl, we can calculate the grams of HCl needed using the molar mass of HCl:
The molar mass of HCl is the sum of the atomic masses of hydrogen (H) and chlorine (Cl):
Molar mass of HCl = 1.01 g/mol (for hydrogen) + 35.45 g/mol (for chlorine)
Molar mass of HCl = 36.46 g/mol
Now, calculate the grams of HCl:
Grams of HCl = (5.00 moles) * (36.46 g/mol)
Grams of HCl = 182.30 grams
So, 2.50 moles of magnesium will consume approximately 182.30 grams of hydrochloric acid in the given reaction.
b. Radiation
c. Dust
d. Shock
Answer:
Se + 2e- => Se-2
Explanation:
Ions are electrically charged particles, formed due to the gain or loss of electrons by an atom.
There are two types of ions cations and anions. An element that loses their electrons and forms positive ions are cations while metals and element that gains one or more electrons and forms negative ions are anions.
Selenium is an atom with atomic number 34 and represented as Se. The electronic configuration of Se is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4. Selenium will gain 2 more electrons to complete it p orbital and form an ion Se-2.
The equation is as below:
Se + 2e- => Se-2