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2 years ago

Which of the following is correct concerning Hess's law?

2 answers:

3 0

I believe the correct answer would be that the change in enthalpy can be found by adding the enthalpies of the individual thermochemical reactions of a chemical reaction. In Hess' Law, enthalpy is independent of the mechanism of the reaction. The enthalpy should be the sum of all the changes in the reaction.

lesantik [10]2 years ago

3 0

(C) the change in enthalpy can be found in adding the enthalpies of the individual thermochemical reactions of a chemical reaction

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Pesticide concentrations in the Rhine River between Germany and France between 1969 and 1975 averaged 0.55 mg/L of hexachloroben

sertanlavr [38]

Answer:

1.93×10⁻³ mmoles/L of C₆Cl₆; 1.58×10⁻⁴ mmoles/L of C₁₂H₈Cl₆O; 3.51×10⁻³ mmoles/L of C₆H₆Cl₆

Explanation:

We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

C₆Cl₆ → 12. 6 + 35.45 .6 = 284.7 g/m

C₁₂H₈Cl₆O → 12 . 12 + 8.1 + 35.45 .6 + 16 = 380.7 g/m

C₆H₆Cl₆ → 12.6 + 6.1 + 35.45 .6 = 290,7 g/m

Let's convert mg to g (/1000)

0.55 mg / 1000 = 5.5×10⁻⁴ g

0.060 mg / 1000 = 6×10⁻⁵ g

1.02 mg / 1000 = 1.02×10⁻³ g

Now we can know the moles (mass / molar mass)

5.5×10⁻⁴ g / 284.7 g/m = 1.93×10⁻⁶ moles of C₆Cl₆

6×10⁻⁵ g / 380.7 g/m = 1.58×10⁻⁷ moles of C₁₂H₈Cl₆O

1.02×10⁻³ g / 290,7 g/m = 3.51×10⁻⁶ moles of C₆H₆Cl₆

Milimoles = Mol . 1000

1.93×10⁻⁶ . 1000 = 1.93×10⁻³ mmoles of C₆Cl₆

1.58×10⁻⁷ . 1000 = 1.58×10⁻⁴ mmoles of C₁₂H₈Cl₆O

3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

6 0

2 years ago

Groups of atoms that are added to carbon backbones and give them unique properties are known as

Irina-Kira [14]

Answer:

Groups of atoms that are added to carbon backbones and give them unique properties are known as Functional Groups.

Explanation:

In organic chemistry they are called as Functional Group because they are the active part of a molecule. These groups give a unique characteristic to molecule both chemically and physically. Also, each functional group represent a different class of compounds.

Examples:

S No. Functional Group Name

1 R--X Alkyl Halides

2 R--OH Alcohols

3 R--NH₂ Amines

4 R--O--R Ethers

5 R--CO--R Ketones

6 R--CO--H Aldehydes

7 R--CO--OH Carboxylic acids

8 R--CO--X Acid Halides

10 R--CO--NR₂ Acid Amides

11 R--CO-OR' Esters

3 0

2 years ago

Ices Creamery combines sugar, cream, and the best natural flavorings, that churns and freezes the resulting mixture at a very lo

Rasek [7]

Answer:

Process manufacturing.

Explanation:

Process manufacturing, a class of manufacturing which is associated with the formulas and the manufacturing of recipes. This type of manufacturing is different from discrete manufacturing as it deals with discrete units and assembly of the components.

This type of manufacturing is common in food, chemical, beverage, nutraceutical, pharmaceutical, cannabis, and mostly biotechnology industries. In this manufacturing, the relevant factors are the ingredients and the bulk materials rather than the individual units.

Thus, formation of ice is an example of process manufacturing in which different ingredients are added and using recipe of freezing.

7 0

2 years ago

A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is

Colt1911 [192]

Answer:

$M_{Na^+}=1.36M$

$M_{Br^-}=1.58M$

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

$n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+$

Once we've got the moles we compute the final volume via:

$V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L$

Thus, the molarity of the sodium atoms turn out into:

$M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M$

Now, we perform the same procedure but now for the bromide ions:

$n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-$

Finally, its molarity results:

$M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M$

Best regards.

7 0

2 years ago

Which compound could serve as a reactant in a neutralization reaction ?

Veseljchak [2.6K]

The compound that could serve as a reactant in the neutralization reaction is H2SO4

Explanation
Neutralization reaction occur between an acids and a base. H2SO4 ( sulfuric acid) is a strong acid. It can be neutralized by strong base such as NaOH ( sodium hydroxide)

Example of neutralization reaction is

2NaOH + H2SO4 → Na2SO4 + 2H2O

7 0

2 years ago