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zlopas [31]
3 years ago
5

If a 45 mM phosphate solution(solution A) had an absorbance of 1.012. What would be the absorbance if 11 mL of solution A was us

ed to prepare a 20.0 mL of a new phosphate solution. Write answer to three significant figures.
Chemistry
1 answer:
timofeeve [1]3 years ago
8 0

Answer:

0.550

Explanation:

The absorbance (A) of a substance depends on its concentration (c) according to Beer-Lambert law.

A = ε . <em>l</em> . c

where,

ε: absorptivity of the species

<em>l</em>: optical path length

A 45 mM phosphate solution (solution A) had an absorbance of 1.012.

A = ε . <em>l</em> . c

1.012 = ε . <em>l</em> . 45 mM

ε . <em>l</em>  = 0.022 mM⁻¹

We can find the concentration of the second solution using the dilution rule.

C₁ . V₁ = C₂ . V₂

45mM . 11mL = C₂ . 20.0 mL

C₂ = 25 mM

The absorbance of the second solution is:

A = (ε . <em>l</em> ). c

A = (0.022 mM⁻¹) . 25 mM = 0.55 (rounding off to 3 significant figures = 0.550)

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3 years ago
When comparing the two chair conformations for a monosubstituted cyclohexane ring, which type of substituent shows the greatest
Aleksandr [31]

Answer:

See the explanation

Explanation:

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In the axial position, we have a more steric hindrance because we have two hydrogens near to the CH_3 group. If we have <u>more steric hindrance</u> the molecule would be <u>more unstable</u>. In the equatorial positions, we don't <u>any interactions</u> because the CH_3 group is pointing out. If we don't have <u>any steric hindrance</u> the molecule will be <u>more stable</u>, that's why the molecule will <u>the equatorial position.</u>

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6 0
3 years ago
Calculate the percentage of the void space out of the total volume occupied by 1 mole of water molecules. The density of water i
gtnhenbr [62]

Answer:

The percentage of the void space out of the total volume occupied is 93.11%

Explanation:

A mole of water contains 2 atoms of hydrogen and 1 atom of oxygen.

To calculate the volume of a mole of water, we calculate 2 times the volume of the hydrogen atom and 1 times the volume of the oxygen atom

Let's calculate this one after the other.

For the hydrogen, formula for the volume will be

V_{hydrogen = 2 × 4/3 × π × r_{H}^{3}

where r_{H}^{3} = 37 pm which is read as 37 picometer (1 picometer = 10^-12 m) = 37 × 10^{-12} meters

Volume of the hydrogen = 8/3 × (37 × 10^-12)^3 = 4.05 * 10^-31

we multiply this by the avogadro's number = 6.02 * 10^23

= 4.05 * 10^-31 * 6.02 * 10^23 = 2.6 * 10^-8 m^3

We do same for thr oxygen, but this time we do not multiply the volume of the oxygen by 2 as we have only one atom of oxygen

Volume of oxygen = 4/3 * π * (73 * 10^-12) ^3 * avogadro's number = 9.81 * 10^-7 m^3

adding both volumes together, we have 1.24 * 10^-6 m^3 or simply 1.24 ml ( 0.01 m = 1 ml)

Dividing the molar mass of one mole of water by its density, we can get the volume of 1 mole of water

= (18g/mol)/(1 g/ml) = 18 ml/mol

Now we proceed to calculate the volume of void = Total volume - volume of molecule = 18 - 1.24 = 16.76 ml

Now, the percentage of void = volume of void/total volume * 100%

= 16.76/18 * 100% = 93.11%

7 0
3 years ago
A solution of 62.4 g of a covalent compound in enough water to make 1.000 L of solution has an osmotic pressure of 0.305 atm at
Alborosie

Answer:

4.99 × 10³ g/mol

Explanation:

Step 1: Given  and required data

  • Mass of the covalent compound (m): 62.4 g
  • Volume of the solution (V): 1.000 L
  • Osmotic pressure (π): 0.305 atm
  • Temperature (T): 25°C = 298 K

Step 2: Calculate the molarity (M) of the solution

The osmotic pressure is a colligative pressure. For a covalent compound, it can be calculated using the following expression.

π = M × R × T

M = π / R × T

M = 0.305 atm / (0.0821 atm.L/mol.K) × 298 K

M = 0.0125 M

Step 3: Calculate the moles of solute (n)

We will use the definition of molarity.

M = n / V

n = M × V

n = 0.0125 mol/L × 1.000 L = 0.0125 mol

Step 4: Calculate the molar mass of the compound

0.0125 moles of the compound weigh 62.4 g. The molar mass is:

62.4 g/0.0125 mol = 4.99 × 10³ g/mol

8 0
3 years ago
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