All categories

3 years ago

What would happen if they had no sunlight???

Chemistry

2 answers:

Sedbober [7]3 years ago

6 0

If they had no sunlight the whole food chain would be thrown off the produce would not grow giving the primary consumer nothing to eat then that and if they have no food they won’t survive and if that were to happen then the secondary consumer would have nothing to eat and it goes on and on for the rest of the chain.

Firdavs [7]3 years ago

4 0

Answer plants will die and with out plants we will die

Explanation:

You might be interested in

A car travelling at 50m/s produces a force of 200N. What is the power of the engine in kilowatts(KW).

Zina [86]

Answer:

10 kilowatts or 10,000 kw

Explanation:

Power = Force × Velocity

Given that:

Velocity of car is 50 m/s

The force is 200N

The power would be:

P = 200 × 50

P = 10,000 W

Notice that it only reaches watts because of the unit of the velocity is m/s. Standard unit to make Kw is km.

Since:

1 kw = 1000 w

10 kw = 10000 w

Credits to the person below me for correction.

6 0

3 years ago

Read 2 more answers

What is the molar solubility in water of ag2cro4 ? (the ksp for ag2cro4 is 8 x 10-12. )

nikitadnepr [17]

The molar solubility in water of Ag₂CrO₄ is 2.52 * 10⁻⁴ M .

<h3>How do you define Solubility ?</h3>

The measure of the degree to which a substance gets dissolved in a solvent to become a solution.

The formula for determination of molar solubity is

$\rm K_{sp} = [A^+]^a [B^-]^b$

$\rm K_{sp}$ = solubility product constant

$\rm [A^+]^a$ = cation in an aqueous solution

$\rm [B^-]^b$ = anion in an aqueous solution

a, b = relative concentrations of a and b

We have the data given as

$\rm K_{sp}$ for Ag₂CrO₄ is 8 x 10⁻¹²

For Silver chromate , the solubility would be only one-half of the Ag⁺ concentration.

We denote the solubility of Ag₂CrO₄ as S mol L⁻¹.

Then for a saturated solution, we have

[Ag⁺]=2S

[CrO₂⁴⁻]=S

the relation between the solubility and the solubility product constant depends on the stoichiometry of the dissolution reaction

(2S)²(S)=4S³ = 8 * 10⁻¹²

S=∛ (2*10⁻¹²)

S= 1.26 * 10⁻⁴ M

2S = 2.52 * 10⁻⁴ M

Therefore the molar solubility in water of Ag₂CrO₄ is 2.52 * 10⁻⁴ M .

To know more about Solubility

brainly.com/question/8591226

#SPJ4

4 0

2 years ago

Read 2 more answers

Eye and hair color are examples of what?

NemiM [27]

the anwser is genotypes which is a personal characsteristic such as skin color eye or hair color or height

3 0

3 years ago

What reagent could be used to separate Br- from CH3 CO2 when added to an aqueous solution containing both.

Serjik [45]

C is the answer for this!

6 0

3 years ago

Read 2 more answers

Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→f

Zanzabum

K = 5.0 × 10^25

<h2>Part (a). Calculate E° for the reaction </h2>

Step 1. Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

Step 2. Identify the cathode and the anode

The half-cell with the more negative E° (Zn) is the anode.

Step 3. Calculate E°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

E° = +0.76 V

<h2>Part (b). Calculate K for the reaction </h2>

The relation between E° and K is

E° = (RT)/(nF)lnK

where

R = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

T = the Kelvin temperature

n = the moles of electrons transferred

F = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]lnK

0.76 = 0.012 85 lnK

lnK = 0.76/0.012 85 = 59.16

K =e^59.16 = 5.0 × 10^25

4 0

3 years ago