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WITCHER [35]
3 years ago
6

How are each of these seeds most likely spread from place to place

Chemistry
1 answer:
NISA [10]3 years ago
6 0
Try this website 
http://sciencelearn.org.nz/Science-Stories/Seeds-Stems-and-Spores/Seed-dispersal
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What is the effect on the equilibrium when sodium formate is added to a solution of formic acid? hcooh( aq) + h ( aq) right arro
artcher [175]

There is no effect on the equilibrium when sodium formate is added to a solution of formic acid for hcooh( aq) + h ( aq) right arrow choo –( aq).

Equilibrium is characterized as a state of equilibrium or a stable situation in which conflicting forces balance one another out and no changes are taking place. In terms of economics, equilibrium occurs when supply and demand are equal. When you are composed and steady, you are in an equilibrium state.

An object is considered to be in an equilibrium condition when all of the forces acting on it are in balance. If the upward forces are equal to the descending forces and the rightward forces are equal to the leftward forces, then the forces are said to be balanced. Several instances of equilibrium include a book that is open and at rest. a vehicle that is going steadily.

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1 year ago
Which change in the h+ ion concentration of an aqueous solution represents a decrease of one unit on the ph scale?
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The answer would be a tenfold increase<span>
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6 0
3 years ago
When ethyl-2-methylpropanoate is treated with sodium ethoxide in ethanol, there is no observable reaction. Explain this observat
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Answer:

No net change in reaction occurs in this nucleophilic acyl subtitution reaction

Explanation:

Sodium ethoxide in ethanol gives nucleophilic acyl substitution reaction with ethyl-2-methylpropanoate.

Here ethoxide group replaces an ethoxide group from ester through addition-ellimination pathway.

So, ultimately, the product of this reaction is identical with reactant i.e. ethyl-2-methylpropanoate is reproduced.

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Mechanistic pathway has been shown below.

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3 years ago
The equilibrium constant for the reaction
Hitman42 [59]

The question is incomplete, here is the complete question:

The equilibrium constant for the reaction

N₂O₄(g)⇌2NO₂ at 2°C is Kc = 2.0

If each yellow sphere represents 1 mol of N₂O₄(g) and each gray sphere 1 mol of NO₂ which of the following 1.0 L containers represents the equilibrium mixture at 2°C?

The image is attached below.

<u>Answer:</u> The system which represents the equilibrium having value of K_c=2.0 is system (b)

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c

For a general chemical reaction:

aA+bB\rightarrow cC+dD

The expression for K_{c} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

For the given chemical equation:

N_2O_4(g)\rightleftharpoons 2NO_2

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}      .......(1)

We are given:

Volume of the container = 1.0 L

Value of K_c = 2.0

Molarity of the substance is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}

For the given images:

  • <u>For a:</u>

Number of Gray spheres = 8 moles

Number of yellow spheres = 4 moles

Putting values in expression 1, we get:

K_c=\frac{(8/1)^2}{(4/1)}\\\\K_c=16

  • <u>For b:</u>

Number of Gray spheres = 4 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

K_c=\frac{(4/1)^2}{(8/1)}\\\\K_c=2

  • <u>For c:</u>

Number of Gray spheres = 6 moles

Number of yellow spheres = 6 moles

Putting values in expression 1, we get:

K_c=\frac{(6/1)^2}{(6/1)}\\\\K_c=6

  • <u>For d:</u>

Number of Gray spheres = 2 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

K_c=\frac{(2/1)^2}{(8/1)}\\\\K_c=\frac{1}{2}

Hence, the system which represents the equilibrium having value of K_c=2.0 is system (b)

3 0
3 years ago
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