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bazaltina [42]
3 years ago
6

You have a freshly prepared 1 M (molar) solution of glucose in water. You carefully pour out a 100 mL sample of that solution. H

ow many glucose molecules are included in that 100 mL sample?
Chemistry
1 answer:
IceJOKER [234]3 years ago
8 0

Answer:

6.022\times 10^{22}glucose molecules are included in that 100 mL sample.

Explanation:

Concentration of freshly prepared glucose solution = 1 M = 1 mol/L

1 L = 1000 ml

This means that 1 mole of glucose is present in 1000 mL of water.

If we have 100 mL of solution. then number of moles of glucose will be L;

\frac{1}{1000}\times 100 mL=0.1 mole

1 mole =  N_A=6.022\times 10^{23} molecules/atoms

Number of molecules of glucose in 0.1 mole :

= 0.1 mol\times 6.022\times 10^{23} molecules=6.022\times 10^{22} moleules

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lana66690 [7]

Answer:

laws

Explanation:

in some areas and countries there is a limit on how many children someone can have, this would clearly regulate population growth.

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8 0
3 years ago
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Within each energy level there are different sublevels that are created. Those electrons that have the lowest
Volgvan
It’s option D cause we know that S orbital is spherical
4 0
3 years ago
The oxygen atom is smaller than the sulfur atom because _____(1) the outer electrons of oxygen are more effectively screened fro
Romashka [77]

Answer: Option (2) is the correct answer.

Explanation:

Atomic number of oxygen atom is 8 and its electronic distribution is 2, 6. So, it contains only 2 orbitals which are closer to the nucleus of the atom.

As a result, the valence electrons are pulled closer by the nucleus of oxygen atom due to which there occurs a decrease in atomic size of the atom.

Whereas atomic number of sulfur is 16 and its electronic distribution is 2, 8, 6. As there are more number of orbitals present in a sulfur atom so, the valence electrons are away from the nucleus of the atom.

Hence, there is less force of attraction between nucleus of sulfur atom and its valence electrons due to which size of sulfur atom is larger than the size of oxygen atom.

Thus, we can conclude that the oxygen atom is smaller than the sulfur atom because the outer orbitals of oxygen are located closer to the nucleus than those of sulfur.

3 0
3 years ago
2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is.
Andrej [43]

2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.  

The reaction of 2-bromo-3,4-dimethylpentane is combined with t-butoxide forms 2 alkene in the elimination reaction due to steric hindrance.  The least stable alkene 3,4 dimethyl - 1- pentene is easy to make. the  t-butoxide is (CH₃)₃CO⁻. The reaction involves in this reaction is E2 elimination reaction. This reaction involves the one step reaction. The product will also form that is 3,4 dimethyl - 2 - pentene.  so the reaction involve Elimination reaction and the product due to steric hindrance is 3,4 dimethyl - 1- pentene

Thus, 2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.  

To learn more about  t-butoxide here

brainly.com/question/12303978

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4 0
1 year ago
Which aqueous solution has the highest boiling point at standard pressure?(1) 1.0 M KC1(aq) (3) 2.0 M KCl(aq)(2) 1.0 M CaC12(aq)
miss Akunina [59]

Answer:

(4) 2.0 M CaCl₂(aq).

Explanation:

  • Adding solute to water elevates the boiling point.
  • The elevation in boiling point (ΔTb) can be calculated using the relation:

<em>ΔTb = i.Kb.m,</em>

where, ΔTb is the elevation in boiling point.

i is the van 't Hoff factor.

  • van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kb is the molal elevation constant of water.

m is the molality of the solution.

<u><em>(1) 1.0 M KCl(aq):</em></u>

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(1.0 m) = 2(Kb).

<u><em>(2) 2.0 M KCl(aq):</em></u>

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(2.0 m) = 4(Kb).

<u><em>(3) 1.0 M CaCl₂(aq):</em></u>

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(1.0 m) = 3(Kb).

<u><em>(4) 2.0 M CaCl₂(aq):</em></u>

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(2.0 m) = 6(Kb).

  • <em>So, the aqueous solution has the highest boiling point at standard pressure is: (4) 2.0 M CaCl₂(aq).</em>

<em></em>

6 0
3 years ago
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