(9.0 x 108) X (3.0 x 10-4)

2. The empirical formula of a molecule is CH2O. In an experiment, the molar mass of the molecule was determined to be 360.3 g/mo

Vanyuwa [196]

Answer:

The answer to your question is:

2.- C₁₂ H₂₄ O₁₂

Explanation:

2.-

Data

CH2O

molar mass = 360.3 g/mol

Molar mass of CH2O = 12 + 2 + 16 = 30g

Divide molar mass given by molar mass obtain

x = 360.3/30

x = 12

Finally

C₁₂ H₂₄ O₁₂

Molar mass = (12 x 12) + (24 x 1) + (16 x 12) = 144 + 24 + 192 = 360 g

3.- First we need to write the complete equation of the reaction and balanced it.

Then, we need to convert the mass given to moles of each compound.

After that, we need used rule of three calculate the amount of products based on the moles of reactants given.

Finally, convert the moles to grams.

4.-

a.- It is a relation between the mass of product obtain in an experiment and the mass of a product obtain theoretically times 100.

b.-

35 g of Mg reacted with excess O2

percent yield = 90%

Actual yield = ?

Formula

Percent yield = (actual yield/theoretical yield) x 100

Equation

2Mg + O2 ⇒ 2MgO

48.62 g of Mg ----------------- 80.62 g of MgO

35g ------------------ x

x = 58 g of MgO (Theoretical yield)

Theoretical yield = 58 g of MgO

Actual yield = percent yield x theoretical yield / 100

= 90 x 58 / 100

= 52. 23 g

5 0

3 years ago

Coal can be used to generate hydrogen gas (a potential fuel) by thefollowing endothermic reaction.C(s) + H2O (g) <==> CO(g

SpyIntel [72]

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

$C(s) + H_2O (g)\leftrightharpoons CO(g) + H_2(g)$

Given that reaction is an endothermic reaction.

For the given options:

a)Adding more C

If the concentration of C that is the reactant is increased, so according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease of concentration of C takes place. Therefore, the equilibrium will shift in the right direction to wards the formation of hydrogen gas.

b) Adding more $H_2O$

If the concentration of water that is the reactant is increased, so according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease of concentration of water takes place. Therefore, the equilibrium will shift in the right direction towards the formation of hydrogen gas.

c) Raising the temperature of the reaction mixture

If the temperature is increased,heat of the equilibrium mixture will also increase so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in heat that is decrease in temperature occurs.

As, this is an endothermic reaction, forward reaction will decrease the temperature. Hence, the equilibrium will shift in the right direction that is towards the formation of hydrogen gas.

d) Increasing the volume of reaction mixture

If the volume of the container is increased, the pressure will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where increase in pressure is taking place. As the number of moles of gas molecules is greater at the product side. So, the equilibrium will shift in the right direction that is towards the formation of hydrogen gas.

e) Adding a catalyst to reaction mixture

Role of catalyst is to attain the equilibrium quickly without disturbing the state of equilibrium. Hence, addition of catalyst will not change the equilibrium of the reaction.

f) Adding an inert gas to reaction mixture

Adding inert gas to the mixture at constant volume will not effect the equilibrium. Hence, addition of an inert gas will not change the equilibrium of the reaction.

6 0

3 years ago

Sound waves are mechanical waves in which the particles in the medium vibrate in a direction parallel to the direction of energy

Marat540 [252]

Longitudinal wave is the another name.

3 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

Identify an alkene and carboxylic acid using primary observations

ehidna [41]

Answer:

The general formula for the carboxylic acids is C nH 2n+1COOH (where n is the number of carbon atoms in the molecule, minus 1).

Explanation:

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4 0

3 years ago