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1 year ago

specify whether each of the items listed is hydrophilic or hydrophobic by dragging the labels into the appropriate box.

Chemistry

1 answer:

erik [133]1 year ago

4 0

Hydrophilic;

Polar compounds

water

ions

glucose

Hydrophobic;

Triglycerides

cholesterol

Lipids

Oil

Carbohydrates

<h3>What are hydrophilic substances?</h3>

The term hydrophilic substances is used to describe the substances that could dissolve in water. They are those substances that are water soluble. Somehow, the substances must be polar because in chemistry like dissolves like.

The hydrophobic substances are those substances that do not dissolve in water. They are nonpolar and interact with other nonpolar substances.

The classification of the substances took place below;

Hydrophilic;

Polar compounds

water

ions

glucose

Hydrophobic;

Triglycerides

cholesterol

Lipids

Oil

Carbohydrates

Learn more about hydrophilic substances:brainly.com/question/4692308

#SPJ1

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Answer : Art conservation requires minute study for the properties of a painting before it undergoes a repair work as it may help the scientist to predict the cause of damage and the areas will be identified and accordingly the damage repair work can be started off. To maintain the overall integrity of the painting intact it needs to be studied well. For repairs to be more discrete and less obstructive it has to be inspected well so that the authenticity of the painting is not lost.

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Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

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3 years ago

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Calculate the atomic mass for copper using the weighted average mass method. Express your answer using two decimal places and in

tatuchka [14]

Answer:

63. 55 amu

Explanation:

Copper is known to exist in two different isotopes which are Cu-63 and Cu-65.

Cu-63 has an atomic mass of 62.93 amu and it has an abundance of 69.15%.

Similarly,

Cu-65 has an atomic mass of 64.93 amu and it has an abundance of 30.85%

Therefore, using the weighted average mass method, the atomic mass of copper is:

Atomic mass of copper = (0.6915*62.93) amu + (0.3085*64.93) amu = 43.52 amu + 20.03 amu = 63.55 amu

Thus, the atomic mass of copper (express in two decimal places) is 63.55 amu

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