Given the following equation and heat of reaction: 2 moles of A combine with 2 moles of B to produce 6 moles of C and 26 kJ What

Answer:

0.1714 (w/w) %

Explanation:

Utilizando la ecuación:

16H+(aq) + 2Cr2O72−(aq) + C2H5OH(aq) → 4Cr3+(aq) + 2CO2(g) + 11H2O(l)

Se emplean 2 moles de ion dicromato (Cr₂O₇²⁻) para titular 1 mol de alcohol (C₂H₅OH)

35.46mL = 0.03546L de una solución de Cr₂O₇²⁻ a 0.05961M utilizada para alcanzar el punto de equivalencia en la titulación contiene:

0.03546L ₓ (0.05961 moles Cr₂O₇²⁻ / L) = 2.114x10⁻³ moles Cr₂O₇²⁻

Dado que 2 moles de dicromato reaccionan por cada mol de alcohol, los moles de alcohol en la muestra de plasma son:

2.114x10⁻³ moles Cr₂O₇²⁻ ₓ ( 1 mol C₂H₅OH / 2 moles Cr₂O₇²⁻) = 1.0569x10⁻³ moles de C₂H₅OH

Como la masa molar del alcohol es 46.07g/mol, la masa de alcohol es:

1.0569x10⁻³ moles de C₂H₅OH ₓ (46.07g / mol) = 0.04869g de C₂H₅OH

Por lo tanto, el porcentaje en masa de alcohol en sangre utilizando los 28.40g de plasma es:

(0.04869g de C₂H₅OH / 28.40g) × 100 = 0.1714 (w/w) %

The reaction will yield 2 mol of H₂O. The balanced chemical equation for this process is: c3h8 (g) + 5o2 (g) → 3co2 (g) + 4h2o (g). Since 5 moles of O₂ create 4 moles of H₂O, 2.5 mol of O₂ will produce: 2.5 x 4/5 = 2 mol of H₂O.

Assuming the edge length of the cube is m.
Since the cube's sides are all equal, you can find its volume using the formula:
volume of cube = m^3

Now, using the provided edge length:
volume = (4.33)^3 = 81.1827 cm^3

According to unit conversion standards, 1 cm^3 = 0.001 liters.
So:
volume of cube = 81.1827 cm^3 = 0.0811827 liters

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.

<pNote, the moles of the acid form remain 0.01 mol as it doesn’t undergo reaction!

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:

Joe correctly mixed the solution.

Explanation:

When evaluating both procedures, it's evident that both Jennifer and Joe weighed the same amount of potassium phosphate, which isn’t the variable here.

The difference is that Jennifer added the solid to 1.0 liters of water, resulting in a final volume greater than 1.0 L, thus her concentration will be lower than 1.0 M.

Joe's solution has a final volume of 1.0 L, which is why his preparation is accurate.