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3 years ago

"Working on rigs is dangerous due to explosions and weather"

Chemistry

2 answers:

GarryVolchara [31]3 years ago

5 0

Using oil as energy
Hope this helps!:)

agasfer [191]3 years ago

3 0

This would apply to when you’re using oil energy

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adam wants to investigate the strengths of different acids and alkalis. He has 2 different acid solutions and 2 different alkali

SCORPION-xisa [38]

Answer:

The strength of an acid or alkali depends on the degree of dissociation of the acid or alkali in water. The degree of dissociation measures the percentage of acid molecules that ionise when dissolved in water. He could use universal indicators or litmus paper for this.

Explanation:

(See answer for the explanation)

5 0

3 years ago

Describe professions for chemists

zepelin [54]

Answer:

A chemist searches for new knowledge about chemicals and use it to improve the way we live. He or she may develop products such as synthetic fibers, drugs and cosmetics. Chemists create processes, including all refining and petrochemical processing, that reduce energy use and pollution.

8 0

3 years ago

8. Write the formulas for and give the names of

kobusy [5.1K]

Answer is c...............

6 0

3 years ago

Calculate the mass percent of 0.485 g of H, which reacts with oxygen to form 2.32 g H2O

Nataly [62]

Answer:

Mass % of hydrogen = 20.9 %

Explanation:

Given data:

Mass of hydrogen = 0.485 g

Mass of water = 2.32 g

Mass percent of hydrogen = ?

Solution:

Mass % of hydrogen = mass of hydrogen / mass of water × 100

Mass % of hydrogen = 0.485 g/ 2.32 g × 100

Mass % of hydrogen = 0.209 × 100

Mass % of hydrogen = 20.9 %

4 0

3 years ago

The combustion of 40.10 g of a compound which contains only C, H, Cl and O yields 58.57 g of CO2 and 14.98 g of H2O. Another sam

elena-14-01-66 [18.8K]

Answer:

The empirical formula is C4H5ClO2

Explanation:

Step 1: Data given

Mass of the sample = 40.10 grams

Mass of CO2 produced = 58.57 grams

Mass of H2O produced = 14.98 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Atomic mass C= 12.01 g/mol

Atomic mass O = 16.0 g/mol

Atomic mass H = 1.01 g/mol

In experiment 2, mass = 75.00 grams and 22.06 grams is Cl

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 58.57 grams / 44.01 g/mol

Moles CO2 = 1.33 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol CO2

For 1.33 moles CO2 we have 1.33 moles C

Step 4: Calculate mass C

Mass C = 1.33 grams * 12.01 g/mol

Mass C = 15.97 grams

Step 5: Calculate moles H2O

Moles H2O= 14.98 grams /18.02 g/mol

Moles H2O = 0.831 moles

Step 6: Calculate moles H

For 1mol H2O we have 2 moles H

For 0.831 moles H2O we have 2*0.831 = 1.662 moles H

Step7: Calculate mass H

Mass H = 1.662 moles * 1.01 g/mol

Mass H = 1.68 grams

Step 8: Calculate mass %

%C = (15.97 grams / 40.10) * 100 %

%C = 39.8 %

%H = (1.68 / 40.10 ) *100%

%H = 4.2 %

%Cl = (22.06 / 75.00 ) * 100%

%Cl = 29.4 %

%O = 100 % - 39.8% - 4.2 % - 29.4 %

%O = 26.6 %

Step 9: Calculate moles in compound

We assume the compound has a mass of 100 grams

Mass C = 39.8 grams

MAss H = 4.2 grams

MAss Cl = 29.4 grams

Mass O = 26.6 grams

Moles C = 39.8 grams / 12.01 g/mol

Moles C = 3.314 moles

Moles H = 4.2 moles / 1.01 g/mol

Moles H = 4.158 moles

Moles Cl =29.4 grams / 35.45 g/mol

Moles Cl = 0.829 moles

Moles O = 26.6 grams / 16.0 g/mol

Moles O = 1.663 moles

Step 10: calculate the mol ratio

We divide by the smallest amount of moles

C: 3.314 moles / 0.829 moles = 4

H: 4.158 moles / 0.829 moles = 5

Cl: 0.829 moles /0.829 moles = 1

O: 1.663 moles / 0.829 moles = 2

This means For each Cl atom we have 4 C atoms, 5 H atoms and 2 O atoms

The empirical formula is C4H5ClO2

5 0

3 years ago