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WITCHER [35]
3 years ago
5

What is the percent yield of titanium (IV) oxide, if 10.2 g are formed when 25.7

Chemistry
1 answer:
yawa3891 [41]3 years ago
7 0

Answer:

Percent yield = 91%

Explanation:

Given data:

Percent yield of TiO₂ = ?

Actual yield = 10.2 g

Mass of TiCl₄ react = 25.7 g

Solution:

Chemical equation:

TiCl₄  + O₂   →    TiO₂  + 2Cl₂

Number of moles of TiCl₄:

Number of moles = mass/molar mass

Number of moles = 25.7 g/ 189.7 g/mol

Number of moles = 0.14 mol

Now we will compare the moles of  TiCl₄  and TiO₂.

                       TiCl₄           :            TiO₂

                         1                :              1

                         0.14           :              0.14

Theoretical yield:

Mass = number of moles ×molar mass

Mass = 0.14 mol × 79.86 g/mol

Mass = 11.2 g

Percent yield:

Percent yield = [actual yield / theoretical yield ]  × 100

Percent yield = [10.2 g / 11.2 g ]× 100

Percent yield = 91%

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The new volume of gas Y is 89.38 mL

From the question given above, the following data were obtained:

Initial volume (V₁) = 82.0 mL

Initial temperature (T₁) = 27.0 °C = 27 + 273 = 300 K

Final temperature (T₂) = 54.0 °C = 54 + 273 = 327 K

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Since the pressure is constant, we shall determine the final volume of gas Y using the Charles' law equation as follow:

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the specific heat capacity of water is 1 cal/gc how much heat in calories is released when 25.7 of water is cooled from 85 to 49
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Answer:

The amount of heat that is released is -925.2 cal

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is the amount of heat that a body can receive or release without affecting its molecular structure, that is, it does not change the state (solid, liquid, gaseous).  In other words, sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state.

The equation that allows to calculate heat exchanges is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.

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Replacing:

Q= 1 \frac{cal}{g*C} *25.7 g* (-36 C)

Solving:

Q= -925.2 cal

<u><em>The amount of heat that is released is -925.2 cal</em></u>

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