How quickly must non-frozen ready-to-eat foods be consumed?

A + B → AB:  ✔ combination

AB → A + B: ✔ breakdown

Hydrocarbon + O2 → CO2 + H2O:  <span>✔ oxidation</span>

AB + CD → AD + CB:  <span>✔ exchange</span>

Answer:

7.3

Explanation:

Using the Henderson Hasselbalch equation, one can determine the pH or pOH of a solution via its pKa. Remember, , and pKa = -logKa, where Ka denotes the acid's equilibrium constant.

The Henderson Hasselbalch formula:

In this context, acid X possesses two ionic forms: the carboxyl group and an alternative form. Initially, we have 0.1 mol/L of acid in 100 mL, which gives:

n1 = (0.1 mol/L)×(0.1 L) = 0.01 mol

Upon dissociation, it yields 0.005 mol of the carboxyl form and 0.005 mol of the other form with stoichiometry assumed constant.

Introducing NaOH at a concentration of 0.1 mol/L and 75 mL, the moles of become:

n2 = (0.1 mol/L)×(0.075 L) = 0.0075 mol

Thus, 0.0075 mol of reacts with 0.005 mol of the carboxyl form, leading to 0.0025 mol of , which in turn reacts with 0.005 mol of the alternating group, leaving 0.0025 mol of the latter.

The new solution’s volume is 175 mL, but the concentrations of both forms remain unchanged in volume, so we can utilize the moles in the equation.

Thus, we arrive at:

6.72 = pKa - log 4

pKa - log 4 = 6.72

pKa = 6.72 + log 4

pKa = 6.72 + 0.6

pKa = 7.3

Answer:

Explanation:

AgNO3 + NaCl --> AgCl + NaNO3

Moles

of AgNO3

= molarity * volume

= 1 * 0.01

= 0.01 mol

for NaCl

= 0.01 * 1

= 0.01 mol.

According to stoichiometry, one mole of silver nitrate corresponds to one mole of NaCl reacted. Hence,

Moles of AgCl generated = 0.01 × 1

= 0.01 mol AgCl produced.

Heat gained by the solution as precipitation occurs:

Solution mass = density × volume

= 1 × 20

= 20 g.

Using q = m * Cp * (T2 - T1)

= 20 * 4.18 * (32.6 - 25.0)

= 635 J

The absorbed heat of 635 J indicates the reaction released -635 J

Thus, Delta H = -635 J/0.01 mol

= -63500 J/mol

= -63.5 kJ/mol.