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podryga [215]
3 years ago
6

How many grams of NaCl do you need to prepare 200 ml of a 5M solution? The molecular weight of NaCl is 58.4 g/mol. Smol 0.2L-1mo

l NaCl lib Imol NaClx 8.glol Nacl Mol 58.19 Nac
Chemistry
1 answer:
Butoxors [25]3 years ago
4 0

Answer:

The answer to your question is: 58.4 g of NaCl

Explanation:

Data

Volume = 200 ml = 0.2 l

Concentration = 5M

MW = 58.4 g

mass NaCl = ?

Formula

             Molarity = (# of moles ) / volume

             # of moles = Molarity x volume

             # of moles = 5 x 0.2

             # of moles = 1

              58.4 g ---------------------- 1 mol

              x          ---------------------  1 mol

              x = (1 x 58.4) / 1

             x = 58.4 g of NaCl

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Answer:

V₁  = 374.71  mL

Explanation:

Given data:

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Final temperature = 86°C

Final volume = 456 mL

Solution:

Initial temperature = 22°C (22+273 = 295 k)

Final temperature = 86°C (86+273 = 359 k)

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₁ = V₂T₁ /T₂

V₁  = 456 mL × 295 K / 359 k

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3 0
3 years ago
A scientist measures the speed of sound in a monatomic gas to be 449 m/s at 20∘C. What is the molar mass of this gas?
Ivan

Answer:

The molar mass of the gas is 36.25 g/mol.

Explanation:

  • To solve this problem, we can use the mathematical relation:

ν = \sqrt{3RT/M}

Where, ν is the speed of light in a gas <em>(ν = 449 m/s)</em>,

R is the universal gas constant <em>(R = 8.314 J/mol.K)</em>,

T is the temperature of the gas in Kelvin <em>(T = 20 °C + 273 = 293 K)</em>,

M is the molar mass of the gas in <em>(Kg/mol)</em>.

ν = \sqrt{3RT/M}

(449 m/s) = √ (3(8.314 J/mol.K) (293 K) / M,

<em>by squaring the two sides:</em>

(449 m/s)² = (3 (8.314 J/mol.K) (293 K)) / M,

∴ M = (3 (8.314 J/mol.K) (293 K) / (449 m/s)² = 7308.006 / 201601 = 0.03625 Kg/mol.

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