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dybincka [34]
3 years ago
13

You have 0.14 moles of sodium chloride (NaCl). How many grams do you have? ( 1 mole of NaCl = 58 grams NaCl) *

Chemistry
1 answer:
Step2247 [10]3 years ago
4 0

Answer:

Option D is correct = 8.12 grams of NaCl

Explanation:

Given data:

Moles of sodium chloride = 0.14 mol

Mass of sodium chloride = ?

Solution:

Formula:

Number of moles = mass of NaCl / Molar mass of NaCl

Molar mass of NaCl = 58 g/mol

Now we will put the values in formula.

0.14 mol = Mass of NaCl / 58 g/mol

Mass of NaCl = 0.14 mol  ×  58 g/mol

Mass of NaCl = 8.12 g of NaCl

Thus, 0.14 moles of NaCl contain 8.12 g of NaCl.

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The by-product of the chlorination of an alkane is​  <u>HCl</u>

Explanation:

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  • Chlorination of alkane gives a mixture of different products.
  • When consider mechanism of alkanes chlorination, free radicals are formed during the reaction to keep the continuous reaction.
  • Different alkyl chloride compounds, extended carbon chains compounds and HCl are formed as products in product mixture.
  • Chlorination byproducts, their toxicodynamics and removal from drinking water.
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7 0
3 years ago
How might two lithospheric plates interact with each other?
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Answer:

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7 0
1 year ago
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Calculate the volume of this irregular solid to the nearest cubic centimeter​
EleoNora [17]

Answer:

8 cm3

Explanation:

The volume of this irregular solid will calculated as the difference between the final volume and the initial volume;

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The maximum amounts of lead and copper allowed in drinking water are 0.015 mg/kg for lead and 1.3 mg/kg for copper. Tell the max
GuDViN [60]

Answer:

  • <em>The maximum amount of copper allowed in 100 g of water is </em><u><em>0.00013 g</em></u>

Explanation:

To find the maximum amount of copper (in grams) allowed in 100 g of water use the maximum amount ratio (1.3 mg / kg)  and set a proportion with the unknown amount of copper (x) and the amount of water (100 g):

First, convert 100 g of water to kg: 100 g × 1 kg / 1000 g = 0.1 kg.

Now, set the proportion:

  • 1.3 mg Cu / 1 Kg H₂O = x / 0.1 kg H₂O

Solve for x:

  • x = 0.1 kg H₂O × 1.3 mg Cu / 1 kg H₂O = 0.13 mg Cu

Convert mg to grams:

  • 0.13 mg × 1 g / 1,000 mg = 0.00013 g

Answer: 0.00013 g of copper.

6 0
2 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
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