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Dmitriy789 [7]
3 years ago
7

Gram for gram, fats in food have much more chemical energy than sugar. One component of fat is stearic acid, C18H36O2. When a sa

mple of 1.02 g of stearic acid was burned completely in a bomb calorimeter, the temperature of the calorimeter rose by 4.26oC. The heat capacity of the calorimeter was 9.43 kJ/oC. Calculate the molar heat of combustion of stearic acid in kilojoules per mole.
Chemistry
1 answer:
vodka [1.7K]3 years ago
3 0

Answer:

\mathbf{-1.12 \times 10^4 \ kJ/mol}

Explanation:

Given that:

mass (m) of the sample = 1.02 g

number of moles of stearic acid  =\dfrac{mass}{molar mass}

= \dfrac{1.02 \ g}{284.49 \ g/mol} \\ \\ = 0.000358 \ moles

The change in temp. \Delta T = 4.26^0 \ C

heat capacity of the calorimeter (c) = 9.43 kJ/° C

Thus, heat due to reaction = cΔT

= 9.43 kJ/° C × 4.26° C

= 40.17 kJ

The heat in kJ/mol = \dfrac{40.17 \ kJ}{0.00358 \ mol}

= 11204.23 kJ/mol

= 1.12 × 10⁴ kJ/mol

As a result of  the reaction is exothermic, the heat reaction of the  combustion is:

\mathbf{-1.12 \times 10^4 \ kJ/mol}

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A pycnometer is a precisely weighted vessel that is used for highly accurate density determinations. Suppose that a pycnometer h
dexar [7]

Answer:

5.758  is the density of the metal ingot in grams per cubic centimeter.

Explanation:

1) Mass of pycnometer = M = 27.60 g

Mass of pycnometer with water ,m= 45.65 g

Density of water at 20 °C = d =998.2 kg/m^3

1 kg = 1000 g

1 m^3=10^6 cm^3

998.2 kg/m^3=\frac{998.2 \times 1000 g}{10^6 cm^3}=0.9982 g/cm^3

Mass of water ,m'= m - M = 45.65 g -  27.60 g =18.05 g

Volume of pycnometer = Volume of water present in it = V

Density=\frac{Mass}{Volume}

V=\frac{m'}{d}=\frac{18.05 g}{0.9982 g/cm^3}=18.08 cm^3

2) Mass of metal , water and pycnometer = 56.83 g

Mass of metal,M' =  9.5 g

Mass of water when metal and water are together ,m''= 56.83 g - M'- M

56.83 g - 9.5 g - 27.60 g = 19.7 g

Volume of water when metal and water are together = v

v=\frac{m''}{d}=\frac{19.7 g}{0.9982 g/cm^3}=19.73 cm^3

Density of metal = d'

Volume of metal = v' =\frac{M'}{d'}

Difference in volume will give volume of metal ingot.

v' = v - V

v'=19.73 cm^3-18.08 cm^3=

v'=1.65 cm^3

Since volume cannot be in negative .

Density of the metal =d'

=d'=\frac{M'}{v'}=\frac{9.5 g}{1.65 cm^3}=5.758 g/cm^3

5 0
3 years ago
How many grams of glucose are needed to form 150g of ethanol?
Svetradugi [14.3K]
<span>1 mole glucose gives 2 moles of ethanol

moles of glucose in 2.4 kg = 2400 / 180.18 = 13.320 moles

so moles of ethanol produced = 2* 13.32 = 26.64 moles

weight of ethanol 26.64 * 46.07

=1227.30 gm or 1.23 Kg</span>
6 0
3 years ago
Where in a lake is the benthic zone?
natta225 [31]

Answer:

The benthic zone is the ecological region at the lowest level of a body of water such as an ocean, lake, or stream, including the sediment surface and some sub-surface layers.

Explanation:

8 0
3 years ago
1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation
Rasek [7]

Answer:

Q1: 728.6 J.

Q2:

a) 668.8 J.

b) 0.3495 J/g°C.

Explanation:

<em>Q1: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation:</em>

  • The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(24.9 mL) = 24.9 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 32.2°C - 25.2°C = 7.0°C).

<em>∴ The amount of heat absorbed by water = Q = m.c.ΔT</em> = (24.9 g)(4.18 J/g°C)(7.0°C) = 728.6 J.

<em>Q2:  Calculate the energy change (q) of the surroundings (water) using the enthalpy equation </em>

<em>qwater = m × c × ΔT.  </em>

<em>We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation.</em>

<em></em>

a) First part: the energy change (q) of the surroundings (water):

  • The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(25 mL) = 25 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 31.6°C - 25.2°C = 6.4°C).

<em>∴ The amount of heat absorbed by water = Q = m.c.ΔT</em> = (25 g)(4.18 J/g°C)(6.4°C) = <em>668.8 J.</em>

<em>b) second part:</em>

<em>Q water = Q unknown metal. </em>

<em>Q unknown metal =  - </em>668.8 J. (negative sign due to the heat is released from the metal to the surrounding water).

<em>Q unknown metal =  - </em>668.8 J = m.c.ΔT.

m = 27.776 g, c = ??? J/g°C, ΔT = (final T - initial T = 31.6°C - 100.5°C = - 68.9°C).

<em>- </em>668.8 J = m.c.ΔT = (27.776 g)(c)( - 68.9°C) = - 1914 c.

∴ c = (<em>- </em>668.8)/(- 1914) = 0.3495 J/g°C.

<em></em>

3 0
3 years ago
Examples of conduction
Sonja [21]

Answer:

Conduction: Touching a stove and being burned. Ice cooling down your hand. Boiling water by thrusting a red-hot piece of iron into it.

Explanation:

3 0
3 years ago
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