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3 years ago

The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and CO2:

1 answer:

3 0

To answer your question I will use dimensional analysis, which is used by cancelling out the units. I will also use the balanced equation provided as a conversion factor.

A) First start out with the 0.300 mol of C6H12O6...
0.300 mol C6H12O6 * (2 mol CO2 / 1 mol C6H12O6) = 0.600 mol CO2

*The significant figures (sig figs) at still three, the 2 is a conversion counting number and does not count*

B) First change 2.00 g of C2H5OH to moles of C2H5OH...
The molecular mass of C2H5OH is...
2(12.01 g/mol) + 5(1.008 g/mol) + 16.00 g/mol + 1.008 g/mol = 46.07 g/mol
This can be used as a conversion factor to change grams to moles.
2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) = 0.0434 mol C2H5OH

Second, you can change the moles of C2H5OH to moles of C6H12O6..
0.0434 mol C2H5OH * (1 mol C6H12O6 / 2 mol C6H12O6) = 0.0217 mol C6H12O6

Third, change moles of C6H12O6 to grams...
MM = 6(12.01 g/mol) + 12(1.008 g/mol) + 6(16.00 g/mol) = 180.16 g/mol
0.0217 mol C6H12O6 * (180.16 g C6H12O6 / 1 mol C6H12O6) = 3.91 g C6H12O6

C) Now I am going to put it all into one long dimensional analysis problem.
MM of CO2 = 44.01 g/mol
MM of C2H5OH = 46.07 g/mol

2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) * (2 mol CO2 / 2 mol C2H5OH) * (44.01 g CO2 / 1 mol CO2) = 1.91 g CO2

I hope this helped and I am sorry that I talked to much, I just didn't want to miss anything!

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A 40.2 g sample of a metal heated to 99.3Â°C is placed into a calorimeter containing 120 g of water at 21.8Â°C. The final temper

aliina [53]

Answer:

B) Iron (c=0.45 J/g°C)

Explanation:

Given that:-

Heat gain by water = Heat lost by metal

Thus,

$m_{water}\times C_{water}\times (T_f-T_i)=-m_{metal}\times C_{metal}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,

$m_{water}\times C_{water}\times (T_f-T_i)=m_{metal}\times C_{metal}\times (T_i-T_f)$

For water:

Mass = 120 g

Initial temperature = 21.8 °C

Final temperature = 24.5 °C

Specific heat of water = 4.184 J/g°C

For metal:

Mass = 40.2 g

Initial temperature = 99.3 °C

Final temperature = 24.5 °C

Specific heat of metal = ?

So,

$120\times 4.184\times (24.5-21.8)=40.2\times C_{metal}\times (99.3-24.5)$

$40.2C_{metal}\left(99.3-24.5\right)=120\times \:2.7\times \:4.184$

$40.2C_{metal}\left(99.3-24.5\right)=1355.616$

$C_{metal}=0.45\ J/g^0C$

<u>This value corresponds to iron. Thus answer is B.</u>

3 0

3 years ago

ompare your TLC analysis with the results of the column chromatography. More polar molecules move more slowly through silica tha

Assoli18 [71]

Biphenyl will have a higher R value than the Methyl Orange.

Explanation:

Biphenyl is a aromatic hydrocarbon and it is a nonpolar molecule.

Methyl Orange is a organic compound with a -SO₃⁻Na⁺ polar functional group which will induce a high polarity in the compound.

You may find the chemical structures of both molecules in the attached picture.

Column chromatography, which use as stationary phase silica gel, is a good technique for separation of the Methyl Orange from Biphenyl.

Being a non-polar molecule, Biphenyl will have a higher R value than the Methyl Orange.

To separate them you use a appropriate solvent as eluent, as exemple chloroform, and Biphenyl will elute first from the column and after that, as a separate phase, Methyl Orange will elute thus separating them.

Learn more about:

chromatography

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7 0

3 years ago

2. HOW MUCH HEAT IS REQUIRED TO BE RELEASED WHEN

jarptica [38.1K]

Answer: -112200J

Explanation:

The amount of heat (Q) released from an heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = ?

Mass of water vapour = 30.0g

C = 187 J/ G°C

Φ = (Final temperature - Initial temperature)

= 100°C - 120°C = -20°C

Then apply the formula, Q = MCΦ

Q = 30.0g x 187 J/ G°C x -20°C

Q = -112200J (The negative sign does indicates that heat was released to the surroundings)

Thus, -112200 joules of heat is released when cooling the superheated vapour.

5 0

4 years ago

Enthalpies of reaction calculated from bond energies and from enthalpies of formation are often, but not always, close to each o

jolli1 [7]

The enthalpy change in a reaction is given by-

ΔH°rxn = ∑nΔH°f,products - ∑nΔH°f,reactants

This can be expressed in terms of bond energy as-

ΔH°rxn = BEreactants - BEproducts

Therefore, the calculated bond energy according to the above equation will be-

ΔH°rxn = [ (C-C) + 2(C-O) + 4(C-H) + 2(O-H) ] - [ (C-C) + 2(C-O) + 4(C-H) + 2(O-H) = 0 kJ/mol

<h3>What is enthalpy change?</h3>

Enthalpy change is a measure of the energy emitted or consumed in a reaction. This can be determined using the following equation which involves standard enthalpy of reactant and product formation:

ΔH°rxn = ∑nΔH°f,products - ∑nΔH°f,reactants

<h3>What is bond energy?</h3>

Bond energy is defined as the amount of energy needed to dissociate a mole of molecules into their individual atoms.

Learn more about the Enthalpy Change here:

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2 years ago

The reaction between nitric oxide and oxygen is described by the following chemical equation: Suppose a two-step mechanism is pr

muminat

Answer:

2NO(g) + O2(g) ---> 2NO2(g)

Explanation:

The mechanism for this reaction involves two elementary reactions in which both are bimolecular as shown below;

NO(g) +O2(g) ----> NO2(g) + O(g)

NO(g) + O(g) ----> NO2(g)

Hence overall balanced reaction equation;

2NO(g) + O2(g) ---> 2NO2(g)

7 0

3 years ago