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egoroff_w [7]
2 years ago
8

125 grams of HCl are used. Zn(s) + 2HCl(aq) = ZnCl2(aq) + H2(g). What is the percent yield if 1.2 moles of hydrogen are actually

produced?

Chemistry
1 answer:
Blababa [14]2 years ago
8 0

Answer:

70%

Explanation:

To find the percent yield, you need to (1) convert grams HCl to moles (via molar mass from periodic table), then (2) convert moles HCl to moles H₂ (via mole-to-mole ratio from equation), then (3) calculate the percent yield H₂ (via percent yield equation).

Zn (s) + 2 HCl (aq) --> ZnCl₂ (aq) + 1 H₂ (g)

Molar Mass (HCl): 1.008 g/mole + 35.45 g/mole
Molar Mass (HCl) = 36.458 g/mole

125 g HCl         1 mole HCl         1 mole H₂
---------------  x  ------------------  x  ------------------  = 1.7 moles H₂
                         36.458 g           2 moles HCl

(actual yield / theoretical yield) x 100% = percent yield

theoretical/calculated yield = 1.7 moles H₂
actual yield = 1.2 moles H₂

(1.2 moles H₂ / 1.7 moles H₂) x 100% = 71%

Therefore, the best percent yield of hydrogen produced is 70%.

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An experiment was designed to test the hypothesis that peanuts have more energy than a chip. The experiment determines calorimet
Svet_ta [14]

Answer:

b Different amounts of food samples were used.

Explanation:

The mass of the two samples needs to be the same in order for the test to be accurate.

4 0
1 year ago
Which type of substance ionizes completely and creates hydronium ions when dissolved in water?
maw [93]

Answer:

Strong acid

Explanation:

An acid is a substance that interacts with water to produce excess hydroxonium ions in an aqueous solution.

Hydroxonium ions are formed as a result of the chemical bonding between the oxygen of water molecules and the protons released by the acid due to its ionisation. This makes aqueous solution of acids conduct electricity.

A strong acid is one that ionizes almost completely. Examples are:

1. Hydrochloric acid

2. Tetraoxosulphate (VI) acid

3. Trioxonitrate (V) acid

4. Hydroiodic acid

5. Hydrobromic acid

8 0
3 years ago
Read 2 more answers
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
Suppose of potassium sulfate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of potassium
dimulka [17.4K]

Answer:

This question is incomplete, here's the complete question:

<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>

Explanation:

Reaction :-

K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4

Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol

Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol

Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L

Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L

Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol

Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.

0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+

Final concentration of potassium cation

= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M

8 0
3 years ago
Which of the following provides cell support in an animal cell? A. microtubule B. mitochondria C. lysosomes D. vacuole
Viktor [21]

Answer:

D.

Vacuole

Explanation:

Hope this helps:)

4 0
3 years ago
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