Write the complete balanced equation for the neutralization reaction that occurs when aqueous hydroiodic acid, HI, and sodium hy

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) %

Answer: C. 151 g

Solution: The balanced equation given is:

From this equation, the ratio of moles between all substances is 1:1. We have 117 grams of LIOH and 141 grams of HBr available and need to calculate the theoretical yield of LiBr.

We should convert each reactant’s grams into moles to identify the limiting reagent since the theoretical yield relies on it.

Molar mass for LiOH = 6.94 + 15.999 + 1.008 = 23.947 grams per mole

Molar mass for HBr = 1.008 + 79.904 = 80.912 grams per mole

To find the moles of each reactant, we divide their grams by their respective molar masses.

Moles of LiOH = = 4.89 mol

Moles of HBr = = 1.74 mol

As there are fewer moles of HBr, it is the limiting reactant. With a 1:1 mol ratio between HBr and LiBr, 1.74 moles of LiBr can be produced.

Molar mass of LiBr = 6.94 + 79.904 = 86.844 grams per mole

The mass of LiBr formed = = 151 g LiBr

Based on calculations, the theoretical yield of LiBr is 151 g, hence the correct answer is C.

Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Identifying the given parameters from the question;

Vapor pressure = 94.4 mm of Hg

The reaction for vaporization is expressed as;

C₆H₆(l) ⇄ C₆H₆(g)

The equilibrium in terms of activities can be defined as:

K = a(C₆H₆(g)) / a(C₆H₆(l))

The activity for pure substances equals one:

a(C₆H₆(l)) = 1

For an ideal gas phase, activity is approximated as the ratio of partial pressure to total pressure. Under standard conditions:

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg is the standard pressure

Thus, we find;

K = 94mmHg / 760mmHg = 0.12421

The formula for Gibbs free energy is:

ΔG = - R·T·ln(K)

Here, R represents the gas constant = 8.314472J/molK

Consequently, the ΔG° for the vaporization of benzene is calculated as:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol

The change in Gibbs free energy for the reaction is determined by the difference between the Gibbs free energy of formation of the products and reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = ΔvG° + ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

The reply is: He did not mention any of these topics.

Explanation: Dalton outlined certain postulates for his atomic theory, which are:

1) Matter consists of indivisible atoms.

2) Atoms from different elements combine in fixed proportions to form compounds.

3) The atomic characteristics of a specific element are uniform, including mass. This means all atoms of a given element share the same mass, whereas atoms from different elements exhibit different masses.

4) During a chemical reaction, atoms are not created or destroyed.

5) Atoms of an element are uniform in mass, size, and all other chemical and physical properties.

From these postulates, it is evident that he focused solely on atoms, neglecting subatomic particles or isotopes.

Explanation:

(a)   In a Face-Centered Cubic (FCC) structure, there are 4 atoms present. Hence, the number of atoms in the specified cell is given by:

       Number of atoms/cell =

                                     = 4

The density can be computed as follows:

          Density =

                =

                        = 11.34

Consequently, the density of the specified substance amounts to 11.34 .

(b)   There exists 1 vacancy for every 500 lead (Pb) atoms. As such, the number of Pb atoms occupying four lattice points can be calculated as follows:

            = 125 unit cells

             =

Thus, we arrive at the conclusion that the number of vacancies per gram is .