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
mr Goodwill [35]
3 years ago
11

Given:

Chemistry
1 answer:
bulgar [2K]3 years ago
5 0

Answer:

The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

Explanation:

Given;

CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol

From the combustion reaction above, it can be observed that;

1 mole of methane (CH₄) released 890 kilojoules of energy.

Now, we convert 59.7 grams of methane to moles

CH₄ = 12 + (1x4) = 16 g/mol

59.7 g of CH₄ = \frac{59.7}{16} = 3.73125 \ moles

1 mole of methane (CH₄) released 890 kilojoules of energy

3.73125 moles of methane (CH₄) will release ?

= 3.73125 moles x  -890 kJ/mol

= -3320.81 kJ

Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

You might be interested in
What information is presented inside each square of the periodic table?
Verdich [7]
The symbol, the atomic mass, the number of protons and electrons
8 0
3 years ago
What happened to the weather after mount tambora epurted?
storchak [24]

Answer:

The eruption of Mount Tambora eventually reduced the average global temperature by as much as 3 °C.

Explanation:

The Mount Tambora eruption was the largest and most destructive volcanic event in recorded history, expelling as much as 150 cubic km (roughly 36 cubic miles) of ash, pumice, and other rock, and aerosols—including an estimated 60 megatons of sulfur—into the atmosphere. As that material mixed with atmospheric gases, it prevented substantial amounts of sunlight from reaching Earth’s surface, eventually reducing the average global temperature by as much as 3 °C.

8 0
3 years ago
Calculate the amount of solute needed for 1000 ml of 15% sodium thiosulfate
maksim [4K]
15% =  15 grams of solute in 100 mL solution

15 g --------------- 100 mL
?? ------------------ 1000 mL

1000 x 15 / 100 = 

15000 / 100 => 150 g of solute

hope this helps!
7 0
3 years ago
In a given column of the periodic table the elements have _____.
kakasveta [241]

Answer:

a. similar properties

Explanation:

In a given column of the periodic table, the elements have similar properties.

A column on the periodic table is known as a group or family. The group is a vertical arrangement of elements  on the periodic table.

Elements in the same group have the same number of valence electrons after their group number.

For example, all the elements in group 1 have one valence electron. Those in group 2 have 2 valence electrons.

The valence electrons of an atom determines its chemical properties. So, all elements in the same group have the same chemical properties.

6 0
3 years ago
How many valence electrons does strontium (Sr) have?
Arisa [49]
Two valence electrons
5 0
3 years ago
Other questions:
  • Find the mass of 3.00 mol of acetic acid, C2H4O2.
    10·1 answer
  • Phosphoric acid, which is commonly used as rust inhibitor, food additive and etching agent for dental and orthopedic use, can be
    6·1 answer
  • In the electrolysis of molten fei3, which product forms at the cathode?
    6·1 answer
  • Now imagine you have several of such dipoles, and place them regularly between the plates. For this part of the pre-lab, you can
    8·1 answer
  • The melting and boiling points of a substance are independent of
    8·2 answers
  • What happens to pressure if temperature increases?
    9·2 answers
  • The reaction between sodium and chlorine that forms table salt is shown
    10·1 answer
  • Convert 0.859 mg to cg.<br> Help me plz :(
    9·1 answer
  • CO + 2H2<br> CH3OH<br> How many molecules of CO are needed to form 600g of CH3OH?
    9·1 answer
  • What is neutralisation reaction? why is it named so? give one example.​
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!