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elixir [45]
2 years ago
8

Example of natural selection​

Chemistry
1 answer:
cestrela7 [59]2 years ago
8 0

Answer: Natural selection is the process in nature by which organisms better adapted to their environment tend to survive and reproduce more than those less adapted to their environment. For example, treefrogs are sometimes eaten by snakes and birds.

Explanation:

You might be interested in
The density of H2O2 is 1.407 g/mL, and the density of O2 is 1.428 g/L. How many liters of O2 can be made from 55 mL H2O2
balandron [24]

Explanation:

mass H2O2 = 55 mL(1.407 g/mL) = 80.85 g

molar mass H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol

moles H2O2 = 80.85 g/34.02 g/mol = 2.377 moles H2O2

For each mole of H2O2 you obtain 0.5 mole of O2 (see the equation).

moles O2 = 2.377 moles H2O2 (1 mole O2)/(2 moles H2O2) = 1.188 moles O2

Now, you need the temperature.  If you are at STP (273 K, and 1.00 atm) then 1 mole of an ideal gas at STP has a volume of 22.4 L.  Without temperature you are not really able to continue.  I will assume you are at STP.

Volume O2 = 1.188 moles O2(22.4 L/mole) = 0.0530 L of O2.

which is  53 mL.

8 0
2 years ago
What discovery was made about particles with an accelerator in 1998
kramer

Answer/Explanation:

In June 1998 in Japan a scientist discovered that neutrinos (which is a type of particle) has weight, mass. This was later proven with some very convincing strong evidence.

<u><em>~ LadyBrain</em></u>

8 0
3 years ago
Use the following information to answer the following question.
Irina-Kira [14]

Answer:

For these problems, we need to compare the theoretical yield that we'd get from performing stoichiometry to the actual yield stated in the problem. % yield is the actual yield/theoretical yield x 100%  

Cu + 2 AgNO₠→ Cu(NOâ‚)â‚‚ + 2 Ag ==> each mole of copper yields two moles of silver  

12.7-g Cu x ( 1 mol Cu /63.5-g Cu) x ( 2 mol Ag / 1 mol Cu) x (108-g Ag / 1 mol Ag) = 43.2-g Ag. This is the theoretical yield. Now, since we got 38.1-g Ag our % yield is:  

38.1-g/43.2-g x 100% = 88.2%

Explanation:

5 0
2 years ago
Use the given half reactions to "construct" an electrolytic cell. Zn^2+ + 2 e^---------&gt;Zn E°cell = -0.76 V Cu^2+ + 2 e^-----
Neko [114]

Answer:

See explanation and image attached

Explanation:

The standard cell potential at 298 K is given by;

E°cathode - E°anode

Hence;

E°cell = 0.34 V - (-0.76 V)

E°cell = 0.34 V + 0.76 V

E°cell = 1.1 V

To reduce Zn^2+ to Zn then Zn must be the cathode, hence;

E°cell = (-0.76 V) - 0.34 V

E°cell = -1.1 V

5 0
3 years ago
15.00 grams of Chromium react with 15.00 grams of hydrobromic acid. Calculate the theoretical yield of the reaction. At STP what
s2008m [1.1K]

Answer:

(a) 18.03 g

(b) 2.105 L

(c) 85.15 %

Step-by-step explanation:

We have the masses of two reactants, so this is a<em> limiting reactant problem.  </em>

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

<em>Step 1</em>. <em>Gather all the information</em> in one place with molar masses above the formulas and masses below them.  

M_r:        52.00   80.91       291.71

                2Cr  +  6HBr ⟶ 2CrBr₃ + 3H₂

Mass/g:  15.00    15.00  

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>  

  Moles of Cr = 15.00 × 1/52.00

  Moles of Cr = 0.2885 mol Cr

Moles of HBr = 15.00 × 1/80.91

Moles of HBr = 0.1854 mol HBr ×  

<em>Step 3</em>. Identify the<em> limiting reactant</em>  

Calculate the moles of CrCl₃ we can obtain from each reactant.  

<em>From Cr</em>:

The molar ratio of CrBr₃:Cr is 2 mol CrBr₃:2 mol Cr

Moles of CrBr₃ = 0.2885 × 2/2

Moles of CrBr₃ = 0.2885 mol CrCl₃

<em>From HBr: </em>

The molar ratio of CrBr₃:HBr is 2 mol CrBr₃:6 mol HBr.

Moles of CrBr₃ = 0.1854 × 2/6

Moles of CrBr₃ = 0.061 80 mol CrBr₃

The limiting reactant is HBr because it gives the smaller amount of CrBr₃.

<em>Step 4</em>. Calculate the <em>theoretical yields</em> of CrBr₃ and H₂.

Theoretical yield of CrBr₃ = 0.061 80 × 291.71/1

Theoretical yield of CrBr₃ = 18.03 g CrCl₃

The molar ratio is 3 mol H₂:6 mol HBr

   Theoretical yield of H₂ = 0.1854 × 3/6

   Theoretical yield of H₂ = 0.092 70 mol H₂

<em>Step 5</em>. Calculate the <em>volume of H₂</em> at STP

STP is 1 bar and 0 °C.

The molar volume of a gas at STP is 22.71 L.

Volume = 0.092 70 × 22.71/1

Volume = 2.105 L

<em>Step 6</em>. Calculate the <em>percent yield </em>

       % Yield = actual yield/theoretical yield × 100 %

Actual yield = 15.35 g

       % yield = 15.35/18.03 × 100

       % yield = <em>85.15 % </em>

8 0
3 years ago
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