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Marta_Voda [28]
4 years ago
8

Complete the following road map for converting volume of A to volume of B for a titration of aqueous solution A with aqueous sol

ution B.
a. multiply by the molarity of B
b. multiply by the moles of B per moles of A
c. divide by the molarity of B
d. multiply by the molarity of A
e. divide by the molarity of A
f. multiply by the moles of A per moles of B

1. volume A (L)
2. moles A
3. moles B
4. volume B (L)
Chemistry
1 answer:
rodikova [14]4 years ago
4 0

Answer:

Explanation:

The solution of known concentration is expressed as molarity. Molarity is the mole fraction of solute (i.e. the dissolved substance) per liter of the solution, Molarity is also commonly called molar concentration.

Mathematically;

\mathtt{Molarity = \dfrac{moles \ of \ solute}{ liters \ of \ solution}}

To copy and complete the road map from the given question, we have the following array:

      Volume A (L)

               ↓

d. multiplied by the molarity of A

               ↓

          moles A

               ↓

b. multiplied by the moles of B / moles of A

               ↓

         moles B

               ↓

c. divided by the molarity of B

               ↓

        volume B (L)

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The standard free energy of formation, ΔG∘f, of a substance is the free energy change for the formation of one mole of the subst
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Answer:

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG∘f=−451.0 kJ/mol

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Explanation:

The spontaneity of a reaction  is given by the value of the standard Gibbs free energy of the reaction (ΔG°rxn). The more negative is the ΔG°rxn, the more spontaneous is a reaction.

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ΔG°rxn = ∑np × ΔG°f(products) − ∑nr × ΔG°f(reactants)

By definition, the standard Gibbs free energy of formation of simple substances in their most stable state is zero. That is why, in the reaction of formation of a compound ΔG°rxn = ΔG°f(product).

<em>Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product? </em>

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<em>     </em>Not feasible. ΔG°rxn = ΔG°f(product) > 0.

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    Not feasible. ΔG°rxn = ΔG°f(product) > 0.

    <em>D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol</em>

    Feasible. ΔG°rxn = ΔG°f(product) < 0.

3 0
4 years ago
4.14 Oxygen requirement for growth on glycerol Klebsiella aerogenes is produced from glycerol in aerobic culture with ammonia as
Iteru [2.4K]

Answer:

0.37 g

Explanation:

The molecular weight for Glycerol = 92

Number of Carbon atoms in glycerol (x)  C_3H_8O_3 = 3

Molecular weight of  the Biomass  ( Klebsiella aerogenes )

= CH_{1.73}O_{0.43}N_{0.24}

= \frac{23.97}{0.92}

= 26.1

From the molecular weight of the Biomass, we can deduce the Degree of reduction for the substrate(glycerol denoted as \delta _g) as follows:

= (4×1)+(1×1.73)-(2×0.43)-(3×0.24)

= 4.15

Given that the yield of the Biomass = 0.40 g

However;

C = Yield of Biomass *\frac{Molecular weight of substrate}{Molecular weight of the Biomass}

C = 0.40*\frac{92}{26.1}

C = 1.41 g

Now , the oxygen requirement can be calculated as:

= \frac{1}{4}*(n*S -  C * \delta _{g})

= \frac{1}{4}(3*4.7-1.41*4.15)

= 2.1 g/mol

Hence, we can say that the needed oxygen = 2.1 g/mol of the substrate consumed.

Now converting it to mass terms; we have:

= 2.1*\frac{number of mole of oxygen}{molecular weight of glycerol}

= 2.1 * \frac{16}{92}

= 0.3652 g

≅ 0.37 g

∴ The oxygen requirement for this culture in mass terms = 0.37 g

3 0
3 years ago
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