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valentina_108 [34]
3 years ago
5

Sodium hydroxide is available commercially as a 50.0% by weight aqueous solution. The density of the solution is 1.53 g/mL. Calc

ulate the molarity of this sodium hydroxide solution. Sodium hydroxide is available commercially as a 50.0% by weight aqueous solution. The density of the solution is 1.53 g/mL. Calculate the molarity of this sodium hydroxide solution. 0.450 M 25.0 M 125. M 19.1 M
Chemistry
1 answer:
bulgar [2K]3 years ago
3 0

Answer:

[NaOH] = 19.1 M

Last option.

Explanation:

Sodium hydroxide is available commercially as a 50.0% by weight aqueous solution.

50% by weight means, that in 100 g of solution we have 50 grams of solute.

Molar mass of solute: 40 g/m

Mass / Molar mass = moles

50 g / 40 g/m = 1.25 moles

These are the moles of solute in 100 grams of solution, so we must use density data, so we can find out our volume.

Solution density = Solution mass / Solution volume

1.53 g/mL = 100 g / Solution volume

Solution volume = 100 g / 1.53 g/mL

Solution volume = 65.3 mL

1.25 moles of solute are in 65.3 mL of solution

Molarity = moles /L

Let's convert 65.3 mL in L

65.3 / 1000 = 0.0653 L

1.25 moles /0.0653 L = 19.1 M

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Read 2 more answers
using the Bohr model for hydrogen: energy = hc/wavelength = 2.18 x 10^-18 Joules (1/nf2 - 1/ni2) N=15 to n=5
soldier1979 [14.2K]

Answer:

Energy lost is 7.63×10⁻²⁰J

Explanation:

Hello,

I think what the question is requesting is to calculate the energy difference when an excited electron drops from N = 15 to N = 5

E = hc/λ(1/n₂² - 1/n₁²)

n₁ = 15

n₂ = 5

hc/λ = 2.18×10⁻¹⁸J (according to the data)

E = 2.18×10⁻¹⁸ (1/n₂² - 1/n₁²)

E = 2.18×10⁻¹⁸ (1/15² - 1/5²)

E = 2.18×10⁻¹⁸ ×(-0.035)

E = -7.63×10⁻²⁰J

The energy lost is 7.63×10⁻²⁰J

Note : energy is lost / given off when the excited electron jumps from a higher energy level to a lower energy level

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