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Travka [436]
3 years ago
11

When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing poi

nt of pure . On the other hand, when of ammonium chloride are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for ammonium chloride in . Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits.
Chemistry
1 answer:
Kipish [7]3 years ago
4 0

Answer:

The figures and parameters to answer this question are not clearly stated, so I will answer it in the closest best way I can.

Explanation:

Molality of alanine = mole / weight of solvent( kg)

= (170×1000)/(89×600)

= 3.18 molal

We know ∆ T​​​​​​f​​​​​ = K​​​​f × m

K​​​​​​f​​​​​ = ∆ T​​​​f / molality

= 7.9/3.18 = 2.48 °c.kg.mol-1

Now using the value of cryoscopic constant we calculate can't Hoff factor for Ammonium chloride.

i = ∆ T​​​​​​f /( K​​​​f × molality of NH​​​​​4​​​​Cl)

= (24.7 × 53.5 × 600) /(2.48 × 170 × 1000)

= 1.8

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Answer:

atoms of metallic element.

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3 years ago
If the frequency of a radio station is 88.1 MHz (8.81 × 107 Hz), what is the wavelength of the wave used by this radio station f
Grace [21]

Hello!

If the frequency of a radio station is 88.1 MHz, the wavelength of the wave used by this radio station for its broadcast is 3.403 m

<h2>Why?</h2>

We are going to use the following equation that shows the relation of the frequency of a wave with its wavelength, knowing that radio waves are electromagnetic waves and they travel at the speed of light (299 792 458 m/s):

Wavelength=\frac{Speed_{Light}}{Frequency}= \frac{299 792 458 m/s}{8.81*10^{7} Hz}=3.402865\approx3.403 m

Have a nice day!

5 0
3 years ago
Read 2 more answers
Determine the empirical formula for a compound that contains 48.6%C, 8.2% H, and 43.2% S by mass.
vagabundo [1.1K]

The empirical formula for a compound that contains 48.6%C, 8.2% H, and 43.2% S by mass is C_3H_6S.

<h3>What is the empirical formula?</h3>

An empirical formula tells us the relative ratios of different atoms in a compound.

We need to calculate the number of moles

Number of a mole of carbon =

48.6 g X (\frac{1 mole }{ 12.0107 g}) =4.05 mole

Number of a mole of hydrogen =

8.2g X (\frac{1 mole}{.00784g}) =8.14 mole

Number of moles of sulphur =

43.2g X (\frac{ mole}{32.065g}) = 1.35 mole

Dividing each mole using the smallest number that is divided by 1.35 moles.

Carbon= \frac{4.05 mole }{1.35 mole} =3

Oxygen= \frac{8.14 mole}{1.35 mole} =6

Sulfur= \frac{1.35 mole}{1.35 mole} =1

Empirical formula is C_3H_6S

Learn more about empirical formula here:

brainly.com/question/14044066

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2 years ago
Consider the reaction.
gayaneshka [121]
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3 years ago
3% Hydrogen Peroxide has an oral LD50 of 900 mg/kg. Acetic Acid has an oral LD50 of 3310 mg/kg. Which one is more hazardous to c
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LD50 is defined as the lethal dose 50% which describes the amount of material required to kill 50% of the testing population. It is given in units of mg of chemical per kg of bodyweight of the recipient. 

Comparing hydrogen peroxide and acetic acid, we see that peroxide has a lower LD50 of 900 mg/kg, with acetic acid having LD50 = 3310 mg/kg. When comparing LD50 values, the smaller value will be the more toxic compound. What this means is that in this case, a smaller amount of peroxide is required to kill 50% of the testing population compared to acetic acid.

Therefore, 3% hydrogen peroxide is more hazardous to consume.
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3 years ago
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