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lyudmila [28]
4 years ago
14

Calculate the enthalpy of the reaction

Chemistry
1 answer:
Alik [6]4 years ago
5 0
We write the enthalpy change equation:

2ΔH_formation(NO) = ΔH_reaction + 2ΔH_formation(NO₂)
2(90.2) = ΔHr + 2(33.2)
ΔHr = 114 kJ/mol

The reaction releases 114 kJ/mol

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In your own words, describe in as much detail as possible, what is air made of.
Advocard [28]

Answer:

nitrogen, oxygen, argon, carbon dioxide, neon, helium, krypton, hydrogen, and xenon. It does not include water vapor because the amount of vapor changes based on humidity and temperature.

3 0
3 years ago
Read 2 more answers
The major difference between a 1s orbital and a 2s orbital is that
Neko [114]

Answer:

The 2s orbital is at a higher energy level.

Explanation:

1s and 2s are the sub-orbitals that are located in an atom. They are nearest to the nucleus and are found on the s sub-orbital. The difference between 1s and 2s is the difference in their level of energy. 1s has low energy as compared to 2s. 1s orbital has the lowest energy because it is located closed to the nucleus. 2s orbital has higher energy than 1s because it's orbit is larger than 1s.

5 0
4 years ago
When a capacitor has a potential difference between the plates it is said to be ? . one plate has an excess of electrons and one
Ede4ka [16]

When a capacitor has a potential difference between the plates it is said to be Constant.

Both plates have different charge which signifies that one has higher potential than the other.

Therefore, when we join them in parallel, charge will flow from higher to lower. and it continued to flow until equilibrium (the entire process took only a few seconds), indicating that the potential remains constant.

To learn more about constant potential difference in capacitor visit:

brainly.com/question/3480856

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6 0
1 year ago
Sodium acetate can be formed from the metathesis/double replacement reaction of sodium
telo118 [61]

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.

5 0
3 years ago
You find a compound composed only of element "X" and hydrogen, and know that it is 91.26% element X by mass. Each molecule has 2
worty [1.4K]
Since you know the ratio of atoms, you can start to put a formula togeter. The formula might look like:<span>
X<span>H2.67 
</span></span>but since atoms can't come in fractional amounts, we have to multiply the formula by some number in order to turn 2.67 into a whole #, while still maintaining the ratio. Multiplying 2.67 by 3 yields 8, so the most likely ratio in the molecule is

X3H8<span>so the ratio of 1:2.67 is still maintained. The mass percent tells you that out of every 100g of compound, 91.26g is element X, so the other 8.74g must be H. Dividing each mass by the number of moles in the formula gets us the molar mass of each element (approximately). DIviding 8.74g by 8 gets 1.09, roughly the molar mass of hydrogen. Dividing 91.26g by 3 gets us 30.4, roughly the molar mass of phosphorus. Element X is most likely phosphorus</span>
4 0
3 years ago
Read 2 more answers
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