answer.
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
umka21 [38]
2 years ago
11

1 2 3 4 5 6 7 8 9 10 Suna passes an electric current through a sample of clear, colorless, and odorless liquid. As the experimen

t continues, bubbles form, and the volume of liquid decreases. Suna collects samples of two colorless, odorless gases that bubble out of the liquid. One of the gases burns. Neither the original liquid nor the other gas burns. Which is the best explanation of her results
Chemistry
1 answer:
BaLLatris [955]2 years ago
8 0

The question is incomplete, the complete question is;

Suna passes an electric current through a sample of clear, colorless, and odorless liquid. As the experiment continues, bubbles form, and the volume of liquid decreases. Suna collects samples of two colorless, odorless gases that bubble out of the liquid. One of the gases burns. Neither the original liquid nor the other gas burns. Which is the best explanation of her results? The electric current changed some of the sample to gas even though the sample was not breaking down. Therefore, the original liquid is a compound. The electric current released a gas that was odorless and colorless, like the original sample. Therefore, the original liquid is an element. The sample was broken down by the electric current and formed a new substance that could burn. Therefore, the original liquid is a compound. The sample lost some of its volume, but the gas still had the same chemical makeup as the original sample. Therefore, the original liquid is an element.

Answer:

The sample was broken down by the electric current and formed a new substance that could burn. Therefore, the original liquid is a compound.

Explanation:

When electric current is passed through a compound, the compound may become broken down to release its constituents. We refer to this phenomenon as electrolysis. We can now say that the substance has been 'decomposed' electrolytically.

Since the original sample was decomposed to yield a gas that could burn and one that couldn't burn even though the original sample couldn't burn, then the original sample is a compound.

You might be interested in
If 6.00 g of the unknown compound contained 0.200 mol of C and 0.400 mol of H, how many moles of oxygen, O, were in the sample?
Arada [10]

Convert moles to mass.

mass C = 0.2 mol * 12 g / mol = 2.4 g

mass H = 0.4 mol * 1 g / mol = 0.4 g

So mass left for O = 6 g – (2.4 g + 0.4 g) = 3.2 g

 

Calculating for moles O given mass:

moles O = 3.2 g / (16 g / mol) = 0.2 moles

 

Answer:

<span>0.2 moles O</span>

8 0
2 years ago
Read 2 more answers
The temperature at the boiling point remained constant despite the continued addition of heat by the bunsen burner. What was the
Paul [167]

Concept:

<em><u>Latent Heat of Vaporization</u></em>: It is defined as the amount of heat required to change the state of mater without changing of its temperature.

From the given question, the temperature at the boiling point remained constant despite the continued addition of heat by the Bunsen burner. <em>Actually,</em> this amount of heat is used by water to break the intermolecular bonds between the water molecules in the form of latent heat that converts the liquid state of water into vapor state of water.

Hence, the correct option will be d.<u>The energy was used to break the intermolecular bonds between the water molecules. </u>

3 0
2 years ago
When 64.0 g of methanol (CHOH) is burned, 1454 kJ of energy is produced. What is the heat of combustion for methanol?
andreev551 [17]

 The heat  of combustion  for  methanol   is 727  kj/mol


    <em><u>calculation</u></em>

 calculate the moles  of methanol (CH3OH)

moles = mass/molar  mass

molar mass of methanol =  12 +( 1 x3)  +16 + 1= 32 g /mol

moles is therefore= 64.0 g / 32 g/mol =  2 moles


Heat of combustion  is therefore = 1454 Kj / 2 moles =  727  Kj/mol

6 0
2 years ago
The factor 0.01 corresponds to which prefix? A) milli B) deci C) deka D) centi
BARSIC [14]
The answer is: D) centi
5 0
2 years ago
1. Suppose 1.00 g of NaOH is used to prepare 250 mL of an NaOH solution. Compare the expected molarity of this solution to the a
geniusboy [140]

Answer:

0.1M solution of NaOH

Explanation:

1 mole of NaOH - 40g

? moles - 1 g = 1/40 = 0.025 moles.

Molarity of 1.00g of NaOH in 0.25L (250 mL) = no. of moles/volume

= 0.025/0.25

= 0.1M.

8 0
2 years ago
Other questions:
  • Why does increasing the number of trials increase confidence in the results of the experiment?
    6·2 answers
  • if the fundamental frequency of a certain string is 84 hz, what is the frequency of the second harmonic? 42 hz 84 hz 126 hz 168
    13·2 answers
  • Given that the density of TlCl(s) is 7.00 g/cm3 and that the length of an edge of a unit cell is 385 pm, determine how many form
    6·1 answer
  • Jim takes 45 seconds to walk 180 meters north to a store what is jims meters per second
    9·2 answers
  • Classify these properties of the metal lithium as physical or chemical. Physical Chemical light enough to float on water silvery
    7·2 answers
  • A chemical engineer must report the average volume of a certain pollutant produced by the plants under her supervision. Here are
    10·1 answer
  • In part 2 of the experiment, you will be analyzing a sample of household bleach. A 0.0800 g sample of household bleach is comple
    5·1 answer
  • Which of the following atoms would have the longest de Broglie wavelength, if all have the same velocity?
    10·1 answer
  • For the n = 3 electron shell, which of the following quantum numbers are valid? Check all that apply.
    14·2 answers
  • All amino acids that are found in proteins, except for proline, contain a(n); Group of answer choices amino group carbonyl group
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!