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choli [55]
3 years ago
8

How is the temporory hardness of water remove bye boiling mathod​

Chemistry
1 answer:
IgorLugansk [536]3 years ago
5 0

Answer:

The temporary hardness of water can be removed by boiling. The bicarbonates get converted to insoluble carbonates and settle down at the bottom.

Calcium bicarbonate -------> Calcium carbonate [insoluble] + Water + Carbon dioxide.

Ca[HCO3]2 -----> Ca CO3 + H2O + CO2

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Which energy levels in the atom is the valence shell?
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Which state of matter would demonstrate the best ability to flow/pour?explain using the term particles
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Liquid, in a liquid state of matter, particles will flow over eachother, the forces between the particles are strong enough to hold a certain volume but not strong enough to keep the molecules sliding over each other.


4 0
3 years ago
What elements make up amino acids? Look up the structures of the 20 amino acids in your textbook and list the 5 elements present
borishaifa [10]

Answer:

- carbon (C)

- hydrogen (H)

- oxygen (O)

- nitrogen (N)

- sulfur (S)

Explanation:

Amino acids are organic molecules which base chemical structure is composed by:

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- a carboxyl group (-COOH)

- an amino group (-NH₂)

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According to this, the five chemical elements that are present in amino acids are:

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6 0
3 years ago
Is it true that we always see the same face of the Moon. <br> Yes<br> No Maybe
Hoochie [10]

Answer: no

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4 0
3 years ago
Read 2 more answers
You have a solution with 2.52 moles of isopropanol (C3H8O). The solution weighs 521 grams. What is the percent composition of is
Kamila [148]

Answer:

The Percent composition of isopropanol in the mixture is 29.07 %

Explanation:

Step 1: Data given

Number of moles isopropanol (C3H8O) = 2.52 moles

Mass of the solution = 521 grams

Molar mass of isopropanol (C3H8O) = 60.1 g/mol

Step 2: Calculate mass of isopropanol

Mass isopropanol = moles isopropanol * molar mass isopropanol

Mass isopropanol = 2.52 moles * 60.1 g/mol

Mass isopropanol = 151.45 grams

Step 3: Calculate the percent composition of isopropanol in the mixture

Percent composition of isopropanol = (mass isopropanol / total mass of mixture) * 100 %

Percent composition of isopropanol = (151.45 grams / 521 grams ) * 100 %

Percent composition of isopropanol = 29.07 %

The Percent composition of isopropanol in the mixture is 29.07 %

8 0
3 years ago
Read 2 more answers
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