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denpristay [2]
1 year ago
11

Prepare a journal about the topic Water-The Life Giving Liquid

Chemistry
1 answer:
Alex1 year ago
8 0

Water is known as a life-giving liquid because every living organism depends on the water for its survival.

<h3>Journal about the topic Water-The Life Giving Liquid</h3>

Water is also called a life-giving liquid because without water life on earth is not possible. All other living organisms need to consume water to survive and grow in size. We need water for many other activities such as cleaning, washing, cooking and irrigation. Water is essential for all living things including humans, animals and plants. Water is called wonder liquid as it can dissolve large number of substances. This property enables water to be a great solvent. Most of the chemical reactions of the living world is carried out in water as a medium so that's why water is known as Life Giving Liquid.

So we can conclude that Water is known as a life-giving liquid because every living organism depends on the water for its survival.

Learn more about water here: brainly.com/question/1313076

#SPJ1

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Ethylene produced by fermentation has a specific gravity of 0.787 at 25 degree Celsius. What is the volume of 125g of ethanol at
WITCHER [35]

<u>Answer:</u> The volume of given amount of ethanol at this temperature is 159.44 mL

<u>Explanation:</u>

Specific gravity is given by the formula:

\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}

We are given:

Density of water = 0.997 g/mL

Specific gravity of ethanol = 0.787

Putting values in above equation, we get:

0.787=\frac{\text{Density of a substance}}{0.997g/mL}\\\\\text{Density of a substance}=(0.787\times 0.997g/mL)=0.784g/mL

Density is defined as the ratio of mass and volume of a substance.

\text{Density}=\frac{\text{Mass}}{\text{Volume}} ......(1)

Given values:

Mass of ethanol = 125 g

Density of ethanol = 0.784 g/mL

Putting values in equation 1, we get:

\text{Volume of ethanol}=\frac{125g}{0.784g/mL}=159.44mL

Hence, the volume of given amount of ethanol at this temperature is 159.44 mL

6 0
3 years ago
What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen?
jasenka [17]

Answer:

71.372 g or 0.7 moles

Explanation:

We are given;

  • Moles of Aluminium is 1.40 mol
  • Moles of Oxygen 1.35 mol

We are required to determine the theoretical yield of Aluminium oxide

The equation for the reaction between Aluminium and Oxygen is given by;

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.

Therefore;

1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen

1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium

Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.

4 moles of aluminium reacts to generate 2 moles aluminium oxide.

Therefore;

Mole ratio Al : Al₂O₃ is 4 : 2

Thus;

Moles of Al₂O₃ = Moles of Al × 0.5

                         = 1.4 moles × 0.5

                         = 0.7 moles

But; 1 mole of Al₂O₃ = 101.96 g/mol

Thus;

Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol

                                            = 71.372 g

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True

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