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3 years ago

5. Write word equations for the following reactions

Chemistry

1 answer:

Andrews [41]3 years ago

6 0

Answer:

a) Sulphur + Oxygen → Sulphur dioxide

b) Carbon + Oxygen → Carbon dioxide

c) Sulphur + Iron → Iron sulphide

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Please inbox me I will thanks your all answers

prohojiy [21]

Answer:

Really

Explanation:

3 0

3 years ago

What is the percent yield for the reaction of nitrogen and hydrogen to produce ammonia, if the theoretical yield was 12.5 g but

koban [17]

Answer:

81.6%

Explanation:

10.2/12.5= 0.816

0.816=81.6%

4 0

3 years ago

___________tiny solid particles in interstellar space thought to consist of a core of rocklike material (silicates) or graphite

crimeas [40]

Answer:

Interstellar medium

Explanation:

The interstellar medium (ISM) is the matter and radiation that exists in the space between the star systems in a galaxy. This matter includes gas in ionic, atomic, and molecular form, as well as dust and cosmic rays.

5 0

3 years ago

Draw the line-bond structure of oleic acid (cis-9-octadecenoic acid), CH3(CH2)7CH=CH(CH2)7COOH, at physiological pH. You do not

Elza [17]

Answer:

Explanation:

Oleic acid originates from an unsaturated fatty acid with 18 carbon atoms and one double bond, which can be found in olive oil and many other vegetable and animal oils and fats. Its economic importance includes the production of soap making and cosmetics etc.

The line-bond structure of the given oleic acid in the question can be found in the attached below.

7 0

3 years ago

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose

Slav-nsk [51]

Answer:

The minimum mass of octane that could be left over is 43.0 grams

Explanation:

Step 1: Data given

Mass of octane = 73.0 grams

Mass of oxygen = 105.0 grams

Molar mass octane = 114.23 g/mol

Molar mass oxygen = 32.0 g/mol

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate the number of moles

Moles = mass / molar mass

Moles octane = 73.0 grams / 114.23 g/mol

Moles octane = 0.639 moles

Moles O2 = 105.0 grams / 32.0 g/mol

Moles O2 = 3.28 moles

Step 4: Calculate the limiting reactant

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

O2 is the limiting reactant. It will completely be consumed. (3.28 moles). There will react 3.28 / 12.5 = 0.2624 moles. There will remain 0.639 - 0.2624 = 0.3766 moles octane

Step 5: Calculate mass octane remaining

Mass octane = moles * molar mass

Mass octane = 0.3766 moles * 114.23 g/mol

Mass octane = 43.0 grams

The minimum mass of octane that could be left over is 43.0 grams

3 0

3 years ago