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uranmaximum [27]
3 years ago
7

Vitamin C, also known as ascorbic acid, is water soluble and cannot be produced by the human body. Each day, a person’s diet sho

uld include a source of vitamin C, such as orange juice. Ascorbic acid has a molecular formula of C6H8O6 and a gram-formula mass of 176 grams per mole.
Which formula represents the empirical formula for ascorbic acid?
Chemistry
1 answer:
Brilliant_brown [7]3 years ago
8 0
I wish I could help but I don't understand this problem if u need more help ask ur teacher or look for me help on this site
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Which is a NOT a freshwater source on Earth?
cluponka [151]

Answer:

B the atmosphere it's not on earth and I'm pretty sure  the atmosphere doesn't have water in

Explanation:

7 0
2 years ago
The use of genetics by farmers to produce more abundant and healthier crops is an application of the science of _____.
Klio2033 [76]
<h3>Question :</h3>

The use of genetics by farmers to produce more abundant and healthier crops is an application of the science of _____.

  • DNA profiling

  • biotechnology

  • molecular biology

  • nuclear genetics

<h3>Answer :</h3>

The use of genetics by farmers to produce more abundant and healthier crops is an application of the science of <u>biotechnology</u>.

So correct option is 2nd biotechnology.

<h3>Explanation :</h3>

Biotechnology is the use of biological structures and processes for useful industrial purposes, whether product or function.

5 0
3 years ago
Read 2 more answers
Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32
Law Incorporation [45]

<u>Answer:</u>

<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ

<u>Explanation:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}           ............(1)

  • <u>For a:</u>

The given chemical equation follows:

2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)

<u>Oxidation half reaction:</u>   Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-       ( × 2)

<u>Reduction half reaction:</u>   3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)

We are given:

n=2\\E^o_{cell}=+1.08V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

  • <u>For b:</u>

The given chemical equation follows:

6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)

<u>Oxidation half reaction:</u>   Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-       ( × 6)

<u>Reduction half reaction:</u>   2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)

We are given:

n=6\\E^o_{cell}=-1.29V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0
3 years ago
The reaction of ethane gas (C2H6) with chlorine gas produces C2H5Cl as its main product (along with HCl). In addition, the react
Kazeer [188]

Answer:

The percent yield of  chloro-ethane in the reaction is 82.98%.

Explanation:

C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl

Moles of ethane = \frac{300.0 g}{30 g/mol}=10 mol

Moles of chlorine gases =\frac{650.0 g}{71 .0 g/mol}=9.1549 mol

As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.

This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.

According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.

Then 9.1549 moles of chlorien gas will give:

\frac{1}{1}\times 9.1549 mol=9.1549 mol of chloro-ethane

Mass of 9.1549 moles of chloro-ethane:

9.1549 mol × 64.5 g/mol = 590.4910 g

Theoretical yield of  chloro-ethane: 590.4910 g

Given experimental yield of chloro-ethane: 490.0 g

\% Yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

\%Yield (C_2H_5Cl)=\frac{490.0 g}{590.4910 g}\times 100=82.98\%

The percent yield of  chloro-ethane in the reaction is 82.98%.

6 0
3 years ago
An unknown metal forms a soluble compound, M(NO3)2, which then undergoes electrolysis with an applied current of 2.50 amperes. I
frozen [14]

Answer:

The metal is the cathode

The metal is palladium

At the other cathode, the reaction going on is

M(s) ------> M2^+(aq) + 2e

Chlorine can spontaneously react with Pd converting it to Pd^2+

The molar mass of the metal is 106.4 g of M

Explanation:

Given the formula of the compound, M^2+ ion is involved. The equation of the reaction is;

M^2+(aq) + 2e -------------> M(s)

The number of moles of electrons transferred = 2 moles

1 mole of electron = 96500C

And Q= It

Where I= current

t= time taken in seconds

Now;

x g of the metal (its molar mass) is deposited by 2×96500 C

1.87 g of the metal is deposited by (2.50× 22.6×60) C

x = 1.87 × 2×96500/ 2.50 × 22.6 × 60

x= 360910/3390

x= 106.4 g of M

The metal is the cathode

The metal is palladium

At the other cathode, the reaction going on is

M(s) ------> M2^+(aq) + 2e

Chlorine can spontaneously react with Pd converting it to Pd^2+

7 0
3 years ago
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