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8_murik_8 [283]
3 years ago
8

1 2.2.3 Quiz: Shapes of Molecules

Chemistry
1 answer:
ivann1987 [24]3 years ago
8 0

Answer:

iu

Explanation:

kjk

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Assume the molality of isoborneol in your product is 0.275 mol/kg. What is the melting point of your impure sample given that th
aliina [53]

Answer:

168°C is the melting point of your impure sample.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = T_f = ?

Depression in freezing point = \Delta T_f

Depression in freezing point  is also given by formula:

\Delta T_f=i\times K_f\times m

K_f = The freezing point depression constant

m = molality of the sample  = 0.275 mol/kg

i = van't Hoff factor

We have: K_f = 40°C kg/mol

i = 1 (  non electrolyte)

\Delta T_f=1\times 40^oC kg/mol\times 0.275 mol/kg

\Delta T_f=11^oC

\Delta T_f=T- T_f

T_f=T- \Delta T_f=179^oC-11^oC=168^oC

168°C is the melting point of your impure sample.

4 0
3 years ago
What is the mass percent of nitrogen in:<br><br> 1. N2O<br> 2. NO<br> 3. NO2<br> 4. N2O2
Temka [501]
We’re going to use the mass percent formula shown below:



For the percent by mass N, we’re going to rewrite the equation as:

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Problem Details
Calculate the mass percent composition of nitrogen in each nitrogen-containing compound:

c. NO2
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2 years ago
Which of the following is a true statement about atomic nuclei?
Free_Kalibri [48]

Answer: they have a net positive charge

Explanation:

8 0
3 years ago
What mass in grams of KCl (74.55 g/mol) is produced from 65.8 grams of KClO3 (122.55 g/mol) if the reaction has a yield of 67.2%
marishachu [46]
Y’all what is the answer
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3 years ago
Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine
Yuliya22 [10]

Answer:

  • <u>194 g/mol</u>

Explanation:

<u>1) Content of C:</u>

All the C atoms in the 1.000 mg of caffeine will be found in the 1.813 mg of CO₂.

  • Mass of C in 1.813 mg of CO₂

       Since the atomic mass of C is 12.01 g/mol and the molar of of CO₂ is 44.01, there are 12 mg of C in 44 mg of CO₂ and you can set the proporton:

       12.01 mg C / 44.01 mg CO₂ = x / 1.813 g CO₂

        ⇒ x = 1.813 × 12.01 / 44.01 g of C = 0.49475 mg of C

  • Number of moles of C

      number of moles = mass in g / atomic mass = 0.49475×10⁻³ g / 12.01 g/mol = 4.1195×10⁻⁵ moles = 0.041195 milimol

<u>2) Content of H</u>

All the H atoms in the 1.000 mg of caffeine will be found in the 0.4639 mg of H₂O

  • Mass of H in 0.4639 mg of H₂O

       Since the atomic mass of H is 1.008 g/mol and the molar of of H₂O is 18.015 g/mol, there are 2×1.008 mg of H in 18.015 mg of H₂O and you can set the proporton:

       2×1.008 mg H / 18.015 mg H₂O = x / 0.4639 mg H₂O

        ⇒ x = 0.4639 mg H₂O × 2 × 1.008 mg H / 18.015 mg H₂O = 0.051913 mg H

  • Number of moles of H

      number of moles = mass in g / atomic mass = 0.051913 ×10⁻³ g / 1.008 g/mol = 5.1501× 10⁻⁵ moles = 0.051501 milimol

<u>3) Content of N</u>

All the N atoms in the 1.000 mg of caffeine will be found in the 0.2885 mg of N₂

  • Mass of N in 0.2885 mg of N₂ is 0.2885 mg

  • Number of moles of N

      number of moles = mass in g / atomic mass = 0.2885 ×10⁻³ g / 14.007 g/mol = 2.0597× 10⁻⁵ moles = 0.020597 milimol

<u>4) Content of O</u>

The mass of O is calculated by difference:

  • Mass of O = mass of sample - mass of C - mass of H - mass of N

       Mass of O = 1.000 mg - 0.49475 mg C - 0.051913 mg H - 0.2885 mg N

     Mass of O = 0.1648 mg

  • Moles of O =  0.1648 × 10 ⁻³ g / 15.999 g/mol = 1.0303×10⁻⁵ mol = 0.01030 milimol

<u>5) Ratios</u>

Divide every number of mililmoles by the smallest number of milimoles:

  • C:  0.041195 / 0.01030 = 4
  • H: 0.051501 / 0.01030 = 5
  • N: 0.020597 / 0.01030 = 2
  • O: 0.01030 / 0.01030 = 1

  • C: 4
  • H: 5
  • N: 2
  • O: 1

<u>6) Empirical formula:</u>

  • C₄H₅N₂O₁

<u>7) Calculate the approximate mass of the empirical formula:</u>

  • 4 × 12 + 5 × 1 + 2 × 14 + 1 × 16 =  97 g/mol

So, since that number is not between 150 and 200 g/mol, multiply by 2: 97 × 2 = 194, which is between 150 and 200.

Thus, the estimate is 194 g/mol

7 0
3 years ago
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