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
snow_tiger [21]
3 years ago
10

If you used the method of initial rates to obtain the order for no2, predict what reaction rates you would measure in the beginn

ing of the reaction for initial concentrations of 0.200 m, 0.100 m, & 0.050 m no2.
Chemistry
1 answer:
luda_lava [24]3 years ago
8 0

Answer:

Rate law is: Rate = k[NO2]^2 = 10 x [NO2]^2

When [NO2] = 0.200 M:

Rate = 10 x 0.200^2 = 0.400 M/s

When [NO2] = 0.100 M:

Rate = 10 x 0.100^2 = 0.100 M/s

When [NO2] = 0.050 M:

Rate = 10 x 0.050^2 = 0.0250 M/s

Explanation:

If you used the method of initial rates to obtain the order for no2, predict what reaction rates you would measure in the beginning of the reaction for initial concentrations of 0.200 m, 0.100 m, & 0.050 m no2.

Rate law is: Rate = k[NO2]^2 = 10 x [NO2]^2

When [NO2] = 0.200 M:

Rate = 10 x 0.200^2 = 0.400 M/s

When [NO2] = 0.100 M:

Rate = 10 x 0.100^2 = 0.100 M/s

When [NO2] = 0.050 M:

Rate = 10 x 0.050^2 = 0.0250 M/s

You might be interested in
What pattern appears in the vertical arrangement (columns) of the elements in the Periodic Table, moving from top to bottom?
Grace [21]
Idk if this is the information you need but the energy level gets higher the farther you move down the Periodic Table. Every element in a vertical line has the same energy level.
3 0
3 years ago
2. Calculate the mass of 3.47x1023 gold atoms.
lapo4ka [179]

3.47 x 10^{23} atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x 10^{23}

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X 10^{23}

from the relation,

1 mole of element contains 6.022 x 10^{23} atoms.

so no of moles of gold given = \frac{3.47 X 10^{23}  }{6.022 X 10^{23} }

0.57 moles of gold.

from the relation:

number of moles = \frac{mass}{atomic mass of 1 mole}

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x 10^{23} atoms of gold have mass of 113.44 grams

3 0
3 years ago
HELP ASAP
WARRIOR [948]
The snake,raccoon, and preying mantis
4 0
3 years ago
A balloon moving along a straight path experiences a force of
Angelina_Jolie [31]
The net force is 11.1 because 23.5-12.4 is your net force

hope this helps:)
6 0
3 years ago
What is ionization energy
damaskus [11]
Ionization energy is the energy required to remove an electron from a gaseous atom or ion. The first or initial ionization energy or Ei of an atom or molecule is the energy required to remove one mole of electrons from one mole of isolated gaseous atoms or ions
5 0
3 years ago
Other questions:
  • Chlorine released from chlorofluorocarbons (cfcs) reacts with ozone in the atmosphere to form oxygen. the proposed reactions are
    11·1 answer
  • What is a frame reference
    11·2 answers
  • What are the industrial conditions used in the Haber process and why?
    12·1 answer
  • Under certain conditions, neon (Ne) gas diffuses at a rate of 4.5 centimeters per second. Under the same conditions, an unknown
    14·2 answers
  • The forces between water molecules are stronger than the forces between ethanol molecules. Which liquid would probably be most d
    5·2 answers
  • If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
    13·2 answers
  • What is the [H+] if the pH of a<br> solution is 1.65?
    11·1 answer
  • Question 11 (5 points)
    14·1 answer
  • Is this a scientific model? Use complete sentences to explain why or why not
    8·1 answer
  • How much energy is required to convert 400 g of ice at 0 °C to liquid water at 0 °C?
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!