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1 year ago

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What is the empirical formula of C6H12O6? C2H4O2 CH2O CH2O2 C2H4O

1 answer:

wariber [46]1 year ago

5 0

Answer: The empirical formula for C6H12O6 is CH2O. Every carbohydrate, be it simple or complex, has an empirical formula CH2O

Explanation:

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A water molecule consists of an oxygen atom and two hydrogen atoms covalently bonded in a bent shape. Oxygen however, attracts t

77julia77 [94]

Answer:

D.

Explanation:

Water is polar, for one thing. Polar mixes with polar, nonpolar mixes wih nonpolar. This leaves D.

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2 years ago

Question 1<br><br> Of the following which displays a positive energetic process?

TiliK225 [7]

We can describe a positive energetic process as any process which increases the internal energy of the system.

A positive energetic reaction or process is often referred to as being Endothermic. This means that the system which is performing the process absorbs energy. Some examples include:

Boiling an Egg
Roasting food over a fire (the food is the reference system)

etc

Therefore, we can confirm that a positive energetic process is one in which the system in question absorbs energy, thus increasing its internal energy.

<em>Since I could not locate the options online, I have provided a general explanation of the concept coupled with a few examples.</em>

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2 years ago

True or false? You can follow the progress of a reaction that produces a gas using a sensitive mass

sasho [114]

Answer:

true

Explanation:

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2 years ago

I need help with all of these questions

mafiozo [28]

Number five is decrease

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3 years ago

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

5 0

3 years ago