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
cricket20 [7]
3 years ago
10

What are THREE reasons that a creature might use MIMICRY

Chemistry
1 answer:
8_murik_8 [283]3 years ago
7 0

Answer:

- hide from predators

- in search of food

- shelter and protection

thenks and mark me brainliestt :))

You might be interested in
What is the meaning of blood?
Vanyuwa [196]
Blood is a red coloured fluid which consists of two parts plasma and corpuscles
6 0
3 years ago
50.0g of N2O4 is introduced into an evacuated 2.00L vessel and allowed to come to equilibrium with its decomposition product,N2O
Lady_Fox [76]

The mass of N2O4 in the final equilibrium mixture is 39.45 grams.

The decomposition reaction of N2O4 to 2NO2 can be expressed as:

\mathbf{N_2O_4{(g)} \leftrightarrow 2NO_{2(g)}}

From the parameters given:

  • The mass of N2O4 = 50.0 g
  • The molar mass of N2O4 = 92.011 g/mol

The number of mole of N2O4 can be determined as:

\mathbf{Moles \ of \ N_2O_4 = \dfrac{50.0 g}{92.011 g/mol}}

\mathbf{Moles \ of \ N_2O_4 = 0.5434 \ moles}

  • The volume of the vessel in which N2O4 was evacuated is = 2.0 L

From stochiometry, the concentration of \mathbf{[N2O4] = \dfrac{mole \  of \ N_2O_4}{volume \  of \ N_2O_4}}

\mathbf{[N2O4] = \dfrac{0.5434}{2}}

\mathbf{[N2O4] = 0.2717 \ M}

The I.C.E table can be computed as:

                            \mathbf{N_2O_4{(g)}}           ↔            \mathbf{ 2NO_{2(g)}}

Initial                    0.2717                              0

Change                 -x                                     +2x

Equilibrium           (0.2717 - x)                        2x

The equilibrium constant from the I.C.E table can be expressed as:

\mathbf{K_c = \dfrac{[NO]^2}{[N_2O_4]}}

\mathbf{K_c = \dfrac{(2x)^2}{(0.2717-x)}}

  • Recall that; Kc = 0.133

∴

\mathbf{0.133 = \dfrac{4x^2}{(0.2717-x)}}

0.0361 - 0.133x = 4x²

4x²  + 0.133x - 0.0361 = 0

By solving the above quadratic equation, we have;

x = 0.07978

The Concentration of [NO2] = 2x

  • [NO2] = 2 (0.07978)
  • [NO2] = 0.15956 M

The Concentration of [N2O4] = 0.2717 - x

  • [N2O4] = 0.2717 - 0.07978
  • [NO2] = 0.19192 M

Again, from the decomposition reaction, we can assert that;

  • \mathbf{N_2O_4{(g)} \leftrightarrow 2NO_{2(g)}}

0.19192 M of N2O4 decompose to produce 0.15956 M of NO2.

However, if 5.0 g of NO2 is injected into the vessel, then the number of moles of NO2 injected becomes;

\mathbf{= \dfrac{5.0 \ g}{46 \ g/mol}} \\ \\ \\   \mathbf{= 0.10869\  moles}

The Molarity of NO2 injected now becomes:

\mathbf{= \dfrac{00.10869 }{2} } \\ \\ \\ \mathbf{= 0.05434 \ M }

So, the new moles of [NO2] becomes = 0.15956 + 0.05434

= 0.2139 M

The new I.C.E table can be computed as:

                            \mathbf{N_2O_4{(g)}}           ↔            \mathbf{ 2NO_{2(g)}}

Initial                    0.19192                             0.2139 M

Change                 +x                                     -2x

Equilibrium           (0.19192 + x)                       (0.2139 -2x)

NOTE: The injection of NO2 makes the reaction proceed in the backward direction.

The equilibrium constant from the I.C.E table can be expressed as:

\mathbf{K_c = \dfrac{[NO]^2}{[N_2O_4]}}

\mathbf{K_c = \dfrac{(0.2139 - 2x)^2}{(0.19192+x)}}

  • Recall that; Kc = 0.133

\mathbf{0.133= \dfrac{(0.2139 - 2x)^2}{(0.19192+x)}}

By solving for x;

x = 0.2246 or x = 0.0225

We need to consider the value of x that is less than the initial concentration of NO2(0.2139 M) which is:

x = 0.0225

Now, the final concentration of [N2O4] = (0.19192 + 0.0225)M

= 0.21442 M

The final number of moles of N2O4 = Molarity(concentration) × volume

The final number of moles of N2O4 = (0.21442 × 2) moles

The final number of moles of N2O4 = 0.42884 moles

The mass of N2O4 in the final equilibrium mixture is:

= final number of moles × molar mass of N2O4

= 0.42884 moles × 92 g/mol

= 39.45 grams

Learn more about the decomposition of N2O4 here:

brainly.com/question/25025725

6 0
3 years ago
A. Balance the following equation:
yanalaym [24]

Answer:

1.C+2H2-CH4

Explanation:

when we add two to hydrogen the product is equal to the reactant

7 0
4 years ago
Are isotopes similar to ions, yes or no?
asambeis [7]
No, isotopes are versions of elements, where as ions are atoms or molecules.
8 0
2 years ago
A 125.0 mg sample of an unknown, monoprotic acid was dissolved in 100.0 mL of distilled water and titrated with a 0.050 M soluti
Katarina [22]

Im honestly not really sure ..

i just need points

im sorry

6 0
4 years ago
Other questions:
  • You are out on a hike, and you notice that the line of telephone wires you have been following along a slope is crooked. Two of
    15·1 answer
  • If 2.03 g of oxygen react with carbon monoxide, how many grams of CO2 are formed?
    11·1 answer
  • Which one of the following compounds is an example of a covalent molecule?
    7·1 answer
  • Mutations _______ have a positive effect on an organism
    12·1 answer
  • Phosphodiesterase functions Select one: a. as a membrane-bound receptor molecule. b. to inactivate cyclic AMP. c. to attach phos
    10·1 answer
  • What units do we use to calculate the volume of any substance? (sixth grade level)
    11·1 answer
  • Please help me .........<br>solve it.​
    7·1 answer
  • 1000 cm3 of Helium has a mass of 178.5 grams. What is the density of Helium gas?​
    8·1 answer
  • What is the formula for heat of adsorbtion
    5·1 answer
  • What is the weight of a person on earth? The mass of the person is 71kg
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!