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4 months ago

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If the aluminum block is initially at 25 ∘C, what is the final temperature of the block after the evaporation of the alcohol? As

sume that the heat required for the vaporization of the alcohol comes only from the aluminum block and that the alcohol vaporizes at 25 ∘C. The heat of vaporization of the alcohol at 25 ∘C is 45.4 kJ/mol, the specific heat of aluminum is 0.903 J/g⋅∘C

1 answer:

Alekssandra [3K]4 months ago

7 0

Explanation:

Los datos proporcionados son los siguientes.

Calor de vaporización ( $\Delta H_{vap}$ ) = 45.4 kJ/mol

Capacidad calorífica ( $C_{p}$ ) = 0.903 $j/g^{o}C$

Supongamos que el alcohol usado es $C_{3}H_{8}O$ y su masa es 1.12 g. Además, la masa del bloque de aluminio es de 73.0 g.

Primero, calculamos los moles de alcohol ( $C_{3}H_{8}O$ ) de la siguiente manera.

Número de moles de alcohol = $\frac{mass}{\text{molar mass}}$

= $\frac{1.12 g}{60.1 g/mol}$

= 0.0186 mol

<pPor lo tanto, el calor absorbido por el alcohol ( $q_{alcohol}$ ) es igual al calor perdido por el aluminio ( $q_{aluminium}$ )

$n \times \Delta H_{vap}$ = $-m \times C_{p} \times \Delta T$

$0.0186 mol \times 45.4 kJ/mol = - 73.0 g \times 0.903 J/g^{o}C \times (T_{f} - 25^{o}C)$

12.71 = $- (T_{f} - 25^{o}C)$

$T_{f} = 12.3^{o}C$

Por lo tanto, podemos concluir que la temperatura final del bloque es 12.3^{o}C[/tex].

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A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a

KiRa [2933]

Answer:

There are 5.5668 moles of water for every mole of CuSO₄.

Explanation:

The mass of anhydrous CuSO₄ is:

23.403g - 22.652g = 0.751g.

mass of crucible + lid + CuSO₄ - mass of crucible + lid

Given that the molar mass of CuSO₄ is 159.609g/mol, we calculate the moles:

0.751g × $\frac{1mol}{159,609g}$ = 4.7052x10⁻³ moles CuSO₄

The mass of water in the initial sample is:

23.875g - 0.751g - 22.652g = 0.472g.

mass of crucible + lid + CuSO₄ hydrate - CuSO₄ - mass of crucible + lid

As the molar mass of H₂O is 18.02g/mol, we find the moles:

0.472g × $\frac{1mol}{18,02g}$ = 2.6193x10⁻² moles H₂O

The mole ratio of H₂O to CuSO₄ is:

2.6193x10⁻² moles H₂O / 4.7052x10⁻³ moles CuSO₄ = 5.5668

This indicates there are 5.5668 moles of water per mole of CuSO₄.

I hope this is helpful!

5 0

3 months ago

(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o

Tems11 [2777]

Answer :

The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.

Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100$

Here, Xa is the electronegativity of atom A and Xb is that of atom B.

Given: The compound is TiAl₃.

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 (as shown in the provided image)

Substitute these values into the formula:

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100$

$Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100$

$Percent Ionic character = 1 - e^(^-^0^.^1^) * 100$

The value of e⁻¹ equals 0.90.

Therefore, percent ionic character = (1 - 0.90) × 100

Percent Ionic Character = 10%

Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.

8 0

4 months ago