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AlexFokin [52]
2 years ago
7

1+1 hahahahahhhahahaahahahahahahahahahahahahahaha why u dumb

Chemistry
2 answers:
Contact [7]2 years ago
6 0

1 + 1 = 2

thxs for the points realy tho thxs

-Dominant- [34]2 years ago
4 0

Answer:

1+1=2 Unless this is a trick question. Then it's most likely 11.

Explanation:

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I’ll mark brainliest please
DaniilM [7]
I’m assuming your just writing the formula? If so
Potassium chloride: KCL
Potassium nitride: KNO2
Potassium sulfide: K2S
calcium chloride: CaCl2
Calcium nitride: Ca3N2
Calcium sulfide: CaS
Silver chloride: AgCl
Silver nitride: Ag3N
Silver sulfide: Ag2S
Manganese (||) chloride: MnCl2
Manganese (||) nitride: Mn3N2
Manganese (||) sulfide: MnS
5 0
2 years ago
Electrons orbit the nucleus like planets around the sun.​<br>true or false?
gogolik [260]

Answer: true

Explanation: Electrons orbit the nucleus of an atom the way that planets revolve around the Sun. The electrons are like the planets in the solar system. The sun is like the nucleus in the solar system. The answer to the question is true.

7 0
2 years ago
Read 2 more answers
If a buffer solution is 0.220 M in a weak acid ( Ka=7.4×10−5) and 0.540 M in its conjugate base, what is the pH?
valkas [14]

Answer: the pH of the solution is 4.52

Explanation:

Consider the weak acid as Ha, it is dissociated as expressed below

HA     H⁺  +  A⁻

the Henderson -Haselbach equation can be expressed as;

pH = pKa + log( [A⁻] / [HA])

the weak acid is dissociated into H⁺ and A⁻ ions in the solution.

now the conjugate base of the weak acid HA is

HA(aq) {weak acid}     H⁺(aq)  +  A⁻(aq) {conjugate base}

so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;

pKₐ = -logKₐ

pKₐ = -log ( 7.4×10⁻⁵ )

pKₐ = 4.13

now thw pH is

pH = pKₐ  + log( [A⁻] / [HA])

pH = 4.13 + log( [0.540] / [0.220])

pH = 4.13 + 0.3899

pH = 4.5199 = 4.52

Therefore the pH of the solution is 4.52

6 0
3 years 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
From the graph of Density vs. Concentration, created in Graph 1, what was the relationship between the concentration of the suga
USPshnik [31]

The graph is not given in the question, so, the required graph is attached below:

Answer:

According to the graph, the relationship between the density of the sugar solution and the concentration of the sugar solution is directly proportional to each other as they both are increasing exponentially.

The graph shows that, the density of sugar solution will increase with the increase in concentration of sugar in the solution.

8 0
2 years ago
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