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GalinKa [24]
3 years ago
15

Iron(II) sulfate, an iron supplement , FeSO4 Express your answer to four significant figures and include the appropriate units.

Chemistry
1 answer:
strojnjashka [21]3 years ago
5 0

Answer:

Molar mass = 151.9 g/mol

Explanation:

The molar mass of a compound is obtained by adding u the individual atomic masses in the compound. The unit is g/mol.

In FeSO4, we have one Fe, one S and 4 O.

The atomic masses are given as follows;

Fe = 55.845 u

S =  32.065 u

O = 15.999 u

Molar mass = ( 1 * Fe) + (1 * S) + (4 * O)

Molar mass = (1 * 55.854 ) + ( 1 * 32.065) + (4 * 15.999)

Molar mass = 151.915  g/mol

In four significant figures;

Molar mass = 151.9 g/mol

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How is hygiene and sanitation, such as handwashing, important to stopping the spread of water-borne illnesses? ( 50 POINTS ANSWE
Kamila [148]

Answer: Handwashing. It is estimated that washing hands with soap and water could reduce diarrheal disease-associated deaths by up to 50% 1. Researchers in London estimate that if everyone routinely washed their hands, a million deaths a year could be prevented 2.

Explanation:

6 0
3 years ago
gThe mole fraction of potassium nitrate in an aqueous solution is 0.0194. The solution's density is 1.0627 g/mL. What is the mol
xxMikexx [17]

Answer:

0.595 M

Explanation:

The number of moles of water in 1L = 1000g/18g/mol = 55.6 moles of water.

Mole fraction = number of moles of KNO3/number of moles of KNO3 + number of moles of water

0.0194 = x/x + 55.6

0.0194(x + 55.6) = x

0.0194x + 1.08 = x

x - 0.0194x = 1.08

0.9806x= 1.08

x= 1.08/0.9806

x= 1.1 moles of KNO3

Mole fraction of water= 55.6/1.1 + 55.6 = 0.981

If

xA= mole fraction of solvent

xB= mole fraction of solute

nA= number of moles of solvent

nB = number of moles of solute

MA= molar mass of solvent

MB = molar mass of solute

d= density of solution

Molarity = xBd × 1000/xAMA ×xBMB

Molarity= 0.0194 × 1.0627 × 1000/0.981 × 18 × 0.0194×101

Molarity= 20.6/34.6

Molarity of KNO3= 0.595 M

8 0
3 years ago
Please help!!
antiseptic1488 [7]

Answer:

Explanation:

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4 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
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