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Vaselesa [24]
3 years ago
12

What will be the amount of sugar in milligrams if the size of the milk chocolate bar is reduced from 12.630 g to 4.000 g ?

Chemistry
1 answer:
Ganezh [65]3 years ago
8 0
Whit this information I only can say the amount of sugar in the 4.000 g will be equal to the 31.67 % of the amount of sugar in the 12.630 g of the milk chocolate, assuming the sugar is evenly distributed in the chocolate. I am missing the value of the amount of sugar in the 12.630 g of chocolate to be able to calculate the value in mg 
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The standard emf for the cell using the overall cell reaction below is +2.20 V:
ANEK [815]

Answer:

emf generated by cell is 2.32 V

Explanation:

Oxidation: 2Al-6e^{-}\rightarrow 2Al^{3+}

Reduction: 3I_{2}+6e^{-}\rightarrow 6I^{-}

---------------------------------------------------------------------------------

Overall: 2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}

Nernst equation for this cell reaction at 25^{0}\textrm{C}-

E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}

where n is number of electrons exchanged during cell reaction, E_{cell}^{0} is standard cell emf , E_{cell} is cell emf , [Al^{3+}] is concentration of Al^{3+} and [Cl^{-}] is concentration of Cl^{-}

Plug in all the given values in the above equation -

E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V

So, E_{cell}=2.32V

3 0
2 years ago
How do you build a machine dealing with nucleous
Liono4ka [1.6K]
You use a nucleus with inflection on the arm processor.
5 0
3 years ago
An amino acid is usually more soluble in aqueous solvent at pH extremes than it is at a pH near the isolelectric point of the am
gogolik [260]

Answer:

Option B. At pH extremes, the amino acid molecules mostly carry a net charge, thus increasing their solubility in polar solvent.

C. At very low or very high pH, the amino acid molecules have increased charge, thus form more salt bonds with water solvent molecules.

Explanation:

4 0
2 years ago
The name given to the substance formed when a liquid is miscible in another liquid.
m_a_m_a [10]

Answer:

Solute

Explanation:

5 0
3 years ago
The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma
Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})}  \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}

By replacing the corresponding values we obtain:

\frac{k_1}{k_2}=\frac{exp(-\frac{55000J/mol}{8.314J/molK*673.15K} )}{exp(-\frac{64000J/mol}{8.314J/molK*673.15K} )} =4.8

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.

Best regards.

4 0
3 years ago
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