2.92 x 10^-15/ (5.6 x 10^-3) (4.16 x 10^9)

What is the mass of 80.0mL of ethanol at 20 degress celcius

Oxana [17]

Answer: What is the mass, in grams, of 135 mL of ethanol? d=0.789 g/mL - the ethanol density. V=135 mL - the volume of ethanol. m=0.789g/mL*135mL=106.515g ~ 106.5g- the mass of ethanol.

Explanation:

Hope this helps :)

4 0

3 years ago

How does solubility differ between small versus large carbohydrates? Explain.

EleoNora [17]

Answer with Explanation:

Small and large carbohydrates have diverse structural properties. Such property allow them to have different solubility.

Solubility is a chemical property of a substance that allows it to dissolve in a given solvent.

Carbohydrates are made up of "monosaccharides." These are simple sugars and are considered small carbohydrates. On the contrary, "polysaccharide" is an example of a large carbohydrate. Monosaccharides are soluble in water while many polysaccharides are not soluble in water.

The high solubility of monosaccharides is mainly due to the presence of hydroxide (OH) groups. These groups are always ready to bond with water (H₂O). Many polysaccharides are not soluble in water because of intermolecular interactions that prevents it from binding with water. Its dissolution process is different when compared to the smaller molecules.

So, this explains the answer.

5 0

3 years ago

-QUESTION 5-

vodomira [7]

The tall trees much of the sun
have you ever been in a forest? if you have, you’ve probably noticed that it’s usually very shady, and not a lot of sunlight hits the ground. That’s cause the tall trees are so dense, the sunlight doesn’t reach the ground

5 0

3 years ago

A qué temperatura en grados centígrados se encuentra la Luna, si se sabe que está en 3 Kelvin aproximadamente?

Viktor [21]

Answer:

Can you do it in english please don't understand Spanish much. thanks

4 0

3 years ago

For the following reaction, 7.53 grams of benzene (C6H6) are allowed to react with 8.33 grams of oxygen gas. benzene (C6H6) (l)

juin [17]

Answer:

The maximum amount of CO2 that can be formed is 9.15 grams CO2

O2 is the limiting reactant

There will remain 4.82 grams of benzene

Explanation:

Step 1: Data given

Mass of benzene = 7.53 grams

Mass of oxygen gas = 8.33 grams

Molar mass of benzene = 78.11 g/mol

Molar mass oxygen gas = 32.00 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles C6H6 = 7.53 grams / 78.11 g/mol

Moles C6H6 = 0.0964 moles

Moles O2 = 8.33 grams / 32.00 g/mol

Moles O2 = 0.2603 moles

Step 4: Calculate the limiting reactant

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles

There will remain 0.0964 - 0.0347 = 0.0617 moles benzene

This is 0.0617 moles * 78.11 g/mol = 4.82 grams benzene

Step 5: Calculate moles CO2

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2

Step 6: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.208 moles * 44.01 g/mol

Mass CO2 = 9.15 grams

4 0

4 years ago