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Nadusha1986 [10]
4 years ago
13

The heat of vaporization for ethanol is 0.826 kJ/j. Calculate the heat energy in joules required to boil 45.65g of ethanol

Chemistry
2 answers:
andrey2020 [161]4 years ago
6 0

Answer:

ethanol is already at the ethanol's boiling point: (0.826 kJ/g) x (70.05 g) = 57.8613 kJ = 5.79 x 10^4 J.

Missing: 45.65 ‎| ‎Must include: ‎45.65

Explanation:

Rashid [163]4 years ago
5 0

Answer:

Explanation:

Supposing the temperature of the ethanol is already at the ethanol's boiling point:

(0.826 kJ/g) x (70.05 g) = 57.8613 kJ = 5.79 x 10^4 J

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A reaction was performed in which 4.0 g of cyclohexanol was reacted with an acid catalyst to obtain 2.8 g of cyclohexene. calcul
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Answer:

              %age Yield  =  85.36 %

Solution:

The Balance Chemical Reaction is as follow,

               C₆H₁₂O  +   Acid Catalyst    →    C₆H₁₀  +  Acid Catalyst + H₂O

According to Equation ,

              100 g (1 mole) C₆H₁₂O produces  =  82 g (1 moles) of C₆H₁₀

So,

                       4.0 g of C₆H₁₂O will produce  =  X g of C₆H₁₀

Solving for X,

                      X =  (4.0 g × 82 g) ÷ 100 g

                      X  =  3.28 g of C₆H₁₀   (Theoretical Yield)

As we know,

                     %age Yield  =  (Actual Yield ÷ Theoretical Yield) × 100

                     %age Yield  =  (2.8 g ÷ 3.28 g) × 100

                     %age Yield  =  85.36 %

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After the collapse of a nebular cloud, atoms begin gravitating together to form a condensed center. What happens next in the sta
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Answer:

Depending upon the mass gathered, the next process formation varies:

Nuclear fusion can kick in leading to formation of star. The nuclei fuse to together and energy is liberated in the form of light and heat.

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4 years ago
Read 2 more answers
Freeze-drying is a process used to preserve food. If strawberries are to be freeze-dried, then they would be frozen to -80.00 °C
katrin2010 [14]

Answer:

1. 389 kJ; 2. 7.5 µg; 3. 6.25 days

Explanation:

1. Energy required

The water is converted directly from a solid to a gas (sublimation).

They don't give us the enthalpy of sublimation, but

\Delta_{\text{sub}}H = \Delta_{\text{fus}}H + \Delta_{\text{vap}}H = 6.01 + 40.68 = 46.69 \text{ kJ}\cdot\text{mol}^{-1}

The equation for the process is then

Mᵣ:                         18.02

         46.69 kJ + H₂O(s) ⟶ H₂O(g)

m/g:                       150

(a) Moles of water

\text{Moles} = \text{150 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{8.324 mol}

(b) Heat removed

46.69 kJ will remove 1 mol of ice.

\text{Heat removed} = \text{8.234 mol} \times \dfrac{\text{46.69 kJ}}{\text{1 mol}} = \textbf{389 kJ}\\\text{It takes $\large \boxed{\textbf{389 kJ}}$ to remove 150 g of ice}

2. Mass of water vapour in the freezer

For this calculation, we can use the Ideal Gas Law — pV = nRT

(a) Moles of water

Data:

p = 1.00 \times 10^{-3}\text{ torr } \times \dfrac{\text{1 atm}}{\text{760 torr}} = 1.316 \times 10^{-6}\text{ atm}

V = 5 L

T = (-80 + 273.15) K = 193.15 K

Calculation:

\begin{array}{rcl}pV & = & nRT\\1.316 \times 10^{-6}\text{ atm} \times \text{5 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{193.15 K }\\6.6 \times 10^{-6} & = & 15.85n\text{ mol}^{-1} \\n & = & \dfrac{6.6 \times 10^{-6}}{15.85\text{ mol}^{-1}}\\\\& = & 4.2 \times 10^{-7} \text{ mol}\\\end{array}

(b) Mass of water

\text{Mass} = 4.2 \times 10^{-7} \text{ mol} \times \dfrac{\text{18.02 g}}{\text{1 mol}} = 7.5 \times 10^{-6}\text{ g} = 7.5 \, \mu \text{g}\\\\\text{At any given time, there are $\large \boxed{\textbf{7.5 $\mu$g}}$ of water vapour in the freezer.}

3. Time for removal

You must remove 150 mL of water.

It takes 1 h to remove 1 mL of water.

\text{Time} = \text{150 mL} \times \dfrac{\text{1 h}}{\text{1 mL}} = \text{150 h} = \text{6.25 days}

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4 years ago
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