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Andreas93 [3]
2 years ago
13

How much 1.5 m sodium hydroxide is necessary to exactly neutralize 20.0 ml of 2.5 m phosphoric acid?

Chemistry
1 answer:
andrew-mc [135]2 years ago
8 0
Answer is: 100 mL of <span>sodium hydroxide.
Chemical reaction: 3NaOH + H</span>₃PO₄ → Na₃PO₄ + 3H₂O.
V(H₃PO₄) = 20.0 mL ÷ 1000 mL/L = 0.02 L.
n(H₃PO₄) = V(H₃PO₄) · c(H₃PO₄).
n(H₃PO₄) = 0.02 L · 2.5 mol/L.
n(H₃PO₄) = 0.05 mol.
From chemical reaction: n(H₃PO₄) : n(NaOH) = 1 : 3.
n(NaOH) = 0.15 mol.
V(NaOH) = n(NaOH) ÷ c(NaOH).
V(NaOH) = 0.15 mol ÷ 1.5 mol/L.
V(NaOH) = 0.1 L · 1000 mL/L = 100 mL.
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58.69x (7.5x10¹⁵) / 6.02x10²³

=> 7.31x10⁻⁷ g
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2 years ago
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snow_lady [41]
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1 year ago
Describe the many different forms of energy involved with stretching and releasing a rubber band. What other processes are simil
Nadya [2.5K]

Answer:

Conversion of kinetic energy to potential energy (chemo mechanical energy)

In the state of rest, the rubber is a tangled mass of long chained cross-linked polymer that due to their disorderliness are in a state of increased entropy. By pulling on the polymer, the applied kinetic energy stretches the polymer into straight chains, giving them order and reducing their entropy. The stretched rubber then has energy stored in the form of chemo mechanical energy which is a form of potential energy

Conversion of the stored potential energy in the stretched to kinetic energy

By remaining in a stretched condition, the rubber is in a state of high potential energy, when the force holding the rubber in place is removed, due to the laws of thermodynamics, the polymers in the rubber curls back to their state of "random" tangled mass releasing the stored potential energy in the process and doing work such as moving items placed in the rubber's path of motion such as an object that has weight, w then takes up the kinetic energy 1/2×m×v² which can can result in the flight of the object.

Explanation:

5 0
1 year ago
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Determine whether each description applies to electrophilic aromatic substitution or nucleophilic aromatic substitution.
Alborosie

Answer:

a. electrophilic aromatic substitution

b. nucleophilic aromatic substitution

c. nucleophilic aromatic substitution

d. electrophilic aromatic substitution

e. nucleophilic aromatic substitution

f. electrophilic aromatic substitution

Explanation:

Electrophilic aromatic substitution is a type of chemical reaction where a hydrogen atom or a functional group that is attached to the aromatic ring is replaced by an electrophile. Electrophilic aromatic substitutions can be classified into five classes: 1-Halogenation: is the replacement of one or more hydrogen (H) atoms in an organic compound by a halogen such as, for example, bromine (bromination), chlorine (chlorination), etc; 2- Nitration: the replacement of H with a nitrate group (NO2); 3-Sulfonation: the replacement of H with a bisulfite (SO3H); 4-Friedel-CraftsAlkylation: the replacement of H with an alkyl group (R), and 5-Friedel-Crafts Acylation: the replacement of H with an acyl group (RCO). For example, the Benzene undergoes electrophilic substitution to produce a wide range of chemical compounds (chlorobenzene, nitrobenzene, benzene sulfonic acid, etc).

A nucleophilic aromatic substitution is a type of chemical reaction where an electron-rich nucleophile displaces a leaving group (for example, a halide on the aromatic ring). There are six types of nucleophilic substitution mechanisms: 1-the SNAr (addition-elimination) mechanism, whose name is due to the Hughes-Ingold symbol ''SN' and a unimolecular mechanism; 2-the SN1 reaction that produces diazonium salts 3-the benzyne mechanism that produce highly reactive species (including benzyne) derived from the aromatic ring by the replacement of two substituents; 4-the free radical SRN1 mechanism where a substituent on the aromatic ring is displaced by a nucleophile with the formation of intermediary free radical species; 5-the ANRORC (Addition of the Nucleophile, Ring Opening, and Ring Closure) mechanism, involved in reactions of metal amide nucleophiles and substituted pyrimidines; and 6-the Vicarious nucleophilic substitution, where a nucleophile displaces an H atom on the aromatic ring but without leaving groups (such as, for example, halogen substituents).

3 0
2 years ago
Estimate ΔG°rxn for the following reaction at 387 K. HCN (g) + 2 H2 (g) → CH3NH2 (g) ΔH° = −158.0 kJ; ΔS° = −219.9
Lina20 [59]

Answer:

ΔG°rxn = -72.9 kJ

Explanation:

Let's consider the following reaction.

HCN(g) + 2 H₂(g) → CH₃NH₂(g)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression:

ΔG°rxn = ΔH° - T.ΔS°

where,

ΔH° is the standard enthalpy of the reaction

T is the absolute temperature

ΔS° is the standard entropy of the reaction

ΔG°rxn = -158.0 KJ - 387 K × (-219.9 × 10⁻³ J/K)

ΔG°rxn = -72.9 kJ

4 0
2 years ago
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