A mole equals 6.02 x 10^23 . Answer these questions below.

Chemistry

1 answer:

siniylev [52]2 years ago

7 0

Answer:

1. 1.25 mol ants x 6.02*10^23 ants/1 mol ants = 7.53*10^23 ants

2. 4.92*10^26 pencils x 1 mol pencils/6.02*10^23 pencils = 817 mol pencils

3. 0.26 mol molecules x 6.02*10^23 molecules/1 mol molecules = 1.6*10^23 molecules

4. 3.46*10^19 molecules x 1 mol molecules/6.02*10^23 molecules = 5.75*10^-5 mol molecules

5. 5.3*10^20 atoms x 1 mol atoms/6.02*10^23 atoms = 8.8 mol atoms

6. 0.11 mol atoms x 6.02*10^23 atoms/1 mol atoms = 6.6*10^22 atoms

I would suggest looking into "dimensional analysis" for help with this type of material. Dimensional analysis will stick with you all throughout chemistry, so picking it up will be extremely beneficial.

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Name 3 ores for metals and which metal they yield?

Sphinxa [80]

Iron, aluminum, and silicon.

7 0

3 years ago

When optically active (S)-2-methylcyclopentanone is treated with aqueous base, the compound loses its optical activity. Explain

Sunny_sXe [5.5K]

Answer:

See explanation below

Explanation:

In this case, we need to see which is the structure of this compound. Now, racemization occurs basically because we are in an aqueous basic medium, and the ketone can reacts again with water in the medium to form the starting reagent.

First, the base will take out the Alpha hydrogen from the ketone, then, the negative charge goes down and opens up the carbonile group, forming a double bond in there. Later, with the water of the medium, it reacts and substract a proton, and then, with keto enolic equilibrium, forms again the ketone, but this ketone is different from the start, it will be the R isomer which is not optically active.

See picture below for mechanism

3 0

3 years ago

How many grams of H2O will be formed when 36.8 g H2 is mixed with 40.2 g O2 and allowed to completely react to form water

Artemon [7]

Answer:

45.225 grams of H₂O will be formed when 36.8 g H₂ is mixed with 40.2 g O₂

Explanation:

The balanced reaction is:

2 H₂ + O₂ → 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

H₂: 2 moles
O₂: 1 mole
H₂O: 2 moles

Being the molar mass of the compounds:

H₂: 2 g/mole
O₂: 32 g/mole
H₂O: 18 g/mole

then, by reaction stoichiometry, the following amounts of reactant and product mass participate:

H₂: 2 moles* 2 g/mole= 4 g
O₂: 1 mole* 32 g/mole= 32 g
H₂O: 2 moles* 18 g/mole= 36 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction and a simple rule of three as follows: if by stoichiometry 4 g of H₂ react with 32 g of O₂, 36.8 g of H₂ with how much mass of O₂ will it react?

$mass of O_{2} =\frac{36.8 grams of H_{2}*32 grams of O_{2} }{4 grams of H_{2}}$

mass of O₂=294.4 grams

But 294.4 grams of O₂ are not available, 40.2 grams are available. Since you have less mass than you need to react with 36.8 grams of H₂, oxygen O₂ will be the limiting reagent.

Then you can apply the following rule of three: if by stoichiometry 32 grams of O₂ form 36 grams of H₂O, 40.2 grams of O₂ how much mass of H₂O will it form?

$mass of H_{2}O=\frac{40.2 grams of O_{2} *36 grams of H_{2}O }{32 grams of O_{2} }$