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
ololo11 [35]
3 years ago
12

1)

Chemistry
1 answer:
Delicious77 [7]3 years ago
6 0

Answer:

<u><em></em></u>

  • <u><em>C) How much energy was added to the substance to increase molecule motion? </em></u>

Explanation:

<em>The most relevant question to ask regarding this change</em> must take into account the physical knowledge about matter.

When matter changes from<em> liquid </em>state to <em>gaseous</em> state, a physical change called evaporation, the particles (molecules or atoms) of the <em>pure substance </em>will separate from each other, take up more space and move faster.

<em>Condensation</em> is the opposite to evaporation, thus the option A) is not the most relevant question.

<em>The charge of the particles</em> does not change; so the option B) is not relevant at all.

The particles should gain energy from the surroundings to <em>increase</em> their <em>motion</em> (kinetic energy) when they pass from liquid state, where they move slower, to gas state, where they move faster. Hence, the option<em> C), How much energy was added to the substance to increase molecule motion?</em> , is totally relevant.

Since this is an increase in the <em>kinetic energy of the molecules</em>, the option D) is not relevant.

You might be interested in
Which of the following is not an example of a pure element?<br><br> Ar<br> CO,2<br> O,2
KatRina [158]
O,2
hopefully that helps lol if u have any more questions like that don't be afaid to ask me
7 0
3 years ago
When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were
uysha [10]

Answer:

\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}

(b) Mass of H

\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}

(c)Mass of Fe

(i)In 0.4131g of X

\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}

(ii) In 1.2383 g of X

\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}

5 0
3 years ago
#3. How many liters are in 244 g of Krypton (Kr) gas (83.80 g/mol)?
ValentinkaMS [17]

Explanation:

100 liter because it is measured

5 0
3 years ago
Fill In the blank:<br> Compression is the part of the medium where particles are ________.
STALIN [3.7K]
Compressed is the answer! 

I hope this helps. :)
7 0
4 years ago
Determine the concentration of a solution of sodium hydroxide (NaOH) using a 0.500M Potassium Hydrogen Phthalate (KHP) standard
Svetradugi [14.3K]

Answer:

0.425M NaOH assuming the volume of KHP was 25.50mL and the volume of the NaOH solution was 30.0mL

Explanation:

The KHP reacts with NaOH as follows:

KHP + NaOH → KNaP + H₂O

<em>Where 1 mole of KHP reacts per mole of KNaP</em>

<em />

That means, the moles of KHP added to the NaOH solution = Moles NaOH at equivalence point. With the moles of NaOH and the volume in liters we can find the molar concentration of NaOH.

<em>Assuming the volume added of KHP was 25.50mL and the solution of NaOH contains 30.0mL (0.0300L), the concentration of the NaOH is:</em>

<em />

<em>Moles KHP = Moles NaOH:</em>

25.50mL = 0.02550L * (0.500mol / L) = 0.01275 moles KHP = Moles NaOH

<em>Molarity NaOH:</em>

0.01275 moles NaOH / 0.0300L =

<h3>0.425M NaOH assuming the volume of KHP was 25.50mL and the volume of the NaOH solution was 30.0mL</h3>
7 0
3 years ago
Other questions:
  • Some nitrogen for use in synthesizing ammonia is heated slowly, maintaining the external pressure close to the internal pressure
    11·1 answer
  • 1) Which of the following is the smallest unit of a chemio<br> Tallest unit of a chemical element?
    14·1 answer
  • E
    5·1 answer
  • A particular brand of gasoline has a density of 0.737 g/ml at 25 ∘c. how many grams of this gasoline would fill a 14.1 gal tank?
    6·1 answer
  • Clasifica los siguientes elementos en metales, no metales, metaloides , o gases , nobles : flúor , azufre , nitrogeno , cloro ,
    12·2 answers
  • Why does my dogs poop look green?
    15·1 answer
  • An example of a product made from a synthetic material is (help ASAAPP)
    15·1 answer
  • Formula equation is more informative than a word equation why​
    5·1 answer
  • 60 points please help me i will appreciate it!
    8·1 answer
  • How many of these substances are examples of<br> elements?
    5·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!