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oksano4ka [1.4K]
3 years ago
13

Explain why the ionization energy to remove a second electron from potassium is higher than the ionization energy to remove four

electrons from silicon.
Chemistry
1 answer:
never [62]3 years ago
6 0

Explanation:

It is more difficult to remove electrons from the second shell or energy level because of the imbalance between the positive nuclear charge and the remaining electrons.

  • The amount of energy required to remove electrons in ground state of an atom is the ionization energy.
  • The first ionization energy is the energy needed to remove the most loosely bound electron of an atom in the gas phase in ground state.
  • The second energy has a greater nuclear pull as it is closer to the nucleus.
  • Both potassium and silicon have the same number of energy levels.
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Sodium acetate can be formed from the metathesis/double replacement reaction of sodium
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Answer:

Explanation:

Sodium Acetate Trihydrate BP Specifications

Sodium Acetate BP

C2H3NaO2,3H2O

Action and use

Used in solutions for dialysis; excipient.

DEFINITION

Sodium ethanoate trihydrate.

Content

99.0 per cent to 101.0 per cent (dried substance).

CHARACTERS

Appearance

Colourless crystals.

Solubility

Very soluble in water, soluble in ethanol (96 per cent).

IDENTIFICATION

A. 1 ml of solution S (see Tests) gives reaction (b) of acetates.

B. 1 ml of solution S gives reaction (a) of sodium.

C. Loss on drying (As shown in the Relevant Test).

TESTS

Solution S

Dissolve 10.0 g in carbon dioxide-free water prepared from distilled water R and dilute to 100 ml 100 ml with the same solvent.

Appearance of solution

Solution S is clear and colourless.

pH

7.5 to 9.0.

Dilute 5 ml of solution S to 10 ml with carbon dioxide-free water.

Reducing substances

Dissolve 5.0 g in 50 ml of water, then add 5 ml of dilute sulphuric acid and 0.5 ml of 0.002 M potassium permanganate. The pink colour persists for at least 1 h. Prepare a blank in the same manner but without the substance to be examined.

Chlorides

Maximum 200 ppm.

Sulphates

Maximum 200 ppm.

Aluminium

Maximum 0.2 ppm, if intended for use in the manufacture of dialysis solutions.

Arsenic

Maximum 2 ppm, determined on 0.5 g.

Calcium and magnesium

Maximum 50 ppm, calculated as Ca.

Heavy metals

Maximum 10 ppm.

Iron

Maximum 10 ppm, determined on 10 ml of solution S.

Loss on drying

39.0 per cent to 40.5 per cent, determined on 1.000 g by drying in an oven at 130C.

Sodium Acetate FCC Food Grade, US Food Chemical Codex

C2H3NaO2 Formula wt, anhydrous 82.03

C2H3NaO2·3H2O Formula wt, trihydrate 136.08

DESCRIPTION

Sodium Acetate occurs as colorless, transparent crystals or as a granular, crystalline or white powder. The anhydrous form is hygroscopic; the trihydrate effloresces in warm, dry air. One gram of the anhydrous form dissolves in about 2 mL of water; 1 g of the trihydrate dissolves in about 0.8 mL of water and in about 19 mL of alcohol.

Function: Buffer.

REQUIREMENTS

Identification: A 1:20 aqueous solution gives positive tests for Sodium and for Acetate.

Assay: Not less than 99.0% and not more than 101.0% of C2H3NaO2 after drying.

Alkalinity Anhydrous: Not more than 0.2%; Trihydrate: Not more than 0.05%.

Lead: Not more than 2 mg/kg.

Loss on Drying: Anhydrous: Not more than 1.0%; Trihydrate: Between 36.0% and 41.0%.

Potassium Compounds: Passes test.

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Mike mixes two chemicals in a container. The container quickly becomes very hot.
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The chemicals are reactive with one another
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2 years ago
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What is the molar mass of C3N4? (Do NOT round this number.)
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Answer:

92.06 g/mol

Explanation:

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Acetylene, c2h2, has a standard enthalpy of formation, δh° = 226.7 kj/mol, and a standard entropy change for its formation from
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ΔG⁰ = ΔH⁰ - T ΔS⁰

ΔG⁰ : Standard free energy of formation of acetylene

ΔH⁰ : Standard enthalpy of formation (226.7 kJ/mol)

ΔS⁰ : Standard entropy change (58.8 J / K. mol)

T : Temperature 25°C = 298 K (room temperature)

ΔG⁰ = 226.7 - (298 x 58.8 x 10⁻³) = 209.2 kJ /mol
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On the enzyme hexokinase, ATP reacts with glucose to produce glucose 6-phosphate and ADP five orders of magnitude faster than AT
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Answer:

[The rate differential results from induced fit on the enzyme upon binding of glucose. This conformational change excludes water from gaining access to the -phosphate group on ATP, and brings the same group close to the OH group on carbon 6 of glucose]

Explanation:

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