A mothball, composed of naphthalene (c10h8), has a mass of 1.64 g . part a how many naphthalene molecules does it contain?

1 answer:

6 0

Answer is: 7,5·10²¹ <span>naphthalene molecules.
</span>m(C₁₀H₈) = 1,64 g.
n(C₁₀H₈) = m(C₁₀H₈) ÷ M(C₁₀H₈).
n(C₁₀H₈) = 1,64 g ÷ 128,17 g/mol.
n(C₁₀H₈) = 0,0125 mol.
N(C₁₀H₈) = n(C₁₀H₈) · Na.
N(C₁₀H₈) = 0,0125 mol · 6,022·10²³ 1/mol.
N(C₁₀H₈) = 7,5·10²¹.
Na - Avogadro number.

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a. Δ $H^0_{rxn} = -108.0\frac{kJ}{mol}$

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate Δ $H^0_{rxn}$ , for the gas-phase reaction ) using the Hess's Law.

Δ $H^0_{rxn}$ = $E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}$

Given from the question, the table below shows the corresponding Δ $H^0_{t}(kJ/mol)$ for each compound.

Compound $H^0_{t}(kJ/mol)$

Liquid EO -77.4

$CH_4_(g_)$ -74.9

$CO_(g_)$ -110.5

If we incorporate our data into the above previous equation; we have:

Δ $H^0_{rxn}$ = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

= $-108.0 \frac{kJ}{mol}$

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

$T_{initial}$ = 93.0°C &

the enthalpy of vaporization (Δ $H^0_{vap}$ ) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as = $\frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}$