1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Oksanka [162]
2 years ago
14

The atomic mass of an element is measured relative to the mass of *

Chemistry
2 answers:
Neko [114]2 years ago
4 0

Answer:

the number of atoms of the elements contained in 12.00g of carbon-12

Bad White [126]2 years ago
3 0

Answer:

carbon-12. I'm pretty sure

You might be interested in
Molecular polarity has a direct effect on the behavior of molecular compounds. The attractive forces between polar molecules are
Blababa [14]

Answer:

Stronger

Greater

Higher

Explanation:

Molecules are held together by intermolecular forces. These are forces that act between molecules in a particular state matter. Intermolecular forces depend on the nature of the molecule.

For polar molecules, the intermolecular forces are stronger thus it takes more energy to separate them leading to a higher boiling point of polar molecules irrespective of their molecular mass.

8 0
2 years ago
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
Thirty grams of lead oxide and fifteen grams of ammonia react completely to produce solid lead, nitrogen gas, and liquid water.
Kitty [74]

45 g  Thirty grams of lead oxide and fifteen grams of ammonia react completely to produce solid lead, nitrogen gas, and liquid water.

8 0
3 years ago
Read 2 more answers
(c) Polar solutes were separated by hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase. How would
Lelu [443]

Answer:

(c) The retention time would be higher (d) The retention time would be lower.

Explanation:

For the polar solutes which were separated using the hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase, the retention time would be higher if eluent were changed from 80 vol% to 90 vol% acetonitrile in water.

However, for the polar solutes which were separated using the normal-phase chromatography on bare silica with methyl t=butyl ether and 2-propanol solvent, the retention time would be lower if the eluent were changed from 40 vol% to 60 vol% 2-propanol.

5 0
2 years ago
What mass of precipitate forms when 185.5 ml of 0.533 m naoh is added to 627 ml of a solution that contains 15.8 g of aluminum s
Law Incorporation [45]

Answer:

2,57 g of precipitate.

Explanation:

For the reaction:

6 NaOH + Al₂(SO₄)₃ → 2 Al(OH)₃ + 3 Na₂SO₄

The precipitate is Al(OH)₃.

185,5mL of 0,533M NaOH are:

0,1855L × 0,533M = <em>0,0989 moles NaOH</em>

Moles of Al₂(SO₄)₃ are:

15,8g × \frac{1mol}{342,15g} = <em>0,0462 moles Al₂(SO₄)₃</em>

For the total reaction of 0,0989 moles NaOH with Al₂(SO₄)₃ you need:

0,0989moles NaOH × \frac{1molAl_{2}(SO_{4})_{3}}{6 moles NaOH} = <em>0,0165 moles Al₂(SO₄)₃</em>

As you have <em>0,0462 moles Al₂(SO₄)₃ </em>the limiting reactant is NaOH.

0,0989 moles of NaOH produce:

0,0989moles NaOH × \frac{2molAl(OH)_{3}}{6 moles NaOH} = <em>0,0330 moles of Al(OH)₃</em>

These moles are:

0,0330 moles of Al(OH)₃ × (78 g/mol) = <em>2,57 g of Al(OH)₃ ≡ mass of precipitate</em>

<em></em>

I hope it helps!

<em> </em>

3 0
3 years ago
Other questions:
  • Calculate the density of a block of metal with a volume of 12.5 cm3 and mass of 126.0 g​
    10·1 answer
  • What are some physical properties that could have helped to identify a mystery powder?
    13·1 answer
  • Consider the positions of barium (Ba), sulfur (S), silicon (Si), and calcium (Ca) on the periodic table. The atoms of which elem
    14·2 answers
  • How many atoms of carbon are represented in 3CH3CH2O?<br><br> 2<br><br> 4<br><br> 6<br><br> 8
    7·1 answer
  • Which of the following is one of the defining features of a hurricane?
    9·2 answers
  • Can someone double check my work? (Ideal Gas Law Equation)
    7·1 answer
  • 24. All elements found on the left side of the Periodic Table of
    9·2 answers
  • Calculate the number of atoms in 3.21 mole of Mg?
    11·1 answer
  • What is 6.25 in scientific notation?
    7·2 answers
  • Why isn’t there anyone helping for chemistry I need this by literally tomorrow morning before 8:30 ;-;
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!