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liraira [26]
2 years ago
12

Refer to the table . Which plants would most likely be found in the rain forest

Chemistry
1 answer:
OlgaM077 [116]2 years ago
6 0

Answer: I would think 4 because there are plants every were there like literally there is lots of flowers green bushes trees vines there's lots of wonderful beautiful stuff there so I think the number 4

Explanation:

A rainforest orchid. Orchids are very common plants in the tropical rainforest. The Amazon Rainforest itself is home to more than 40,000 plant species! The most common tree in the Amazon Rainforest is the açai here are some of the plants there are there

Bromeliads Plant (Bromeliaceae)

Emergent Plant

Heliconia Flower (Lobster-Claw)

Orchid Plant

Passion flowers (Passiflora spp.)

Lianas

Vines

Rubber Tree (Hevea brasiliensis)

Cacao (Theobroma cacao)

Giant Water Lilies (Victoria amazonica)

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The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)

The balanced two-half reactions will be,

Oxidation half reaction : Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction : Ni^{2+}+2e^-\rightarrow Ni

The expression for reaction quotient will be :

Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

4 0
3 years ago
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Annette [7]

Answer:

6 * 10^-4M

Explanation:

use this equation

[ H + ]  =  10  −  pH

6 0
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