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viva [34]
3 years ago
6

What is adaptation, and what are some examples of it?

Chemistry
1 answer:
Alja [10]3 years ago
6 0
An adaption is a change or process an organism goes through to become better suited for its environment.

Example: A polar bear having white fur in order to blend in in its environment.
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Which of the following pairs of elements is most likely to form an ionic bond?
balu736 [363]

A pair of elements will most likely form an ionic bond if one is a metal and one is a nonmetal. These types of ionic compounds are composed of monatomic cations and anions. ( K, Cl)...

4 0
2 years ago
Read 2 more answers
How many moles of methane are in 20.32 x 10^16 molecules
Marizza181 [45]

Taking into account the definition of avogadro's number, 3.37×10⁻⁷ moles of methane are 20.32×10¹⁶ molecules.

First of all, you have to know that Avogadro's number indicates the number of particles of a substance (usually atoms or molecules) that are in a mole.

Its value is 6.023×10²³ particles per mole and it applies to any substance.

Then you can apply the following rule of three: if 6.023×10²³ molecules are contained in 1 mole of methane, then 20.32×10¹⁶ molecules are contained in how many moles of methane?

amount of moles of methane= (20.32×10¹⁶ molecules × 1 mole)÷ 6.023×10²³ atoms

Solving:

<u><em>amount of moles of methane= 3.37×10⁻⁷ moles</em></u>

Finally, 3.37×10⁻⁷ moles of methane are 20.32×10¹⁶ molecules.

Learn more about Avogadro's Number:

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7 0
1 year ago
The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A
liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

p^{OH}=14-56.17

      =8.823

The volume of the NH_{3} = 40.00 ml  

convert into the liter= 0.040L

The value of the concentrated NH_{3} =0.10 M

The volume of the NH_{4}Cl= 50.00 ml

convert into the liter= 0.050L

 The value of concentrated NH_{4}Cl= 0.10 M

The volume of the H_{2}So_{4}= 30 ml

convert into the liter= 0.030L  

The value of concentrated H_2So_4=0.05 M

Calculating total volume=(0.40+0.050+0.030)

                                       =0.120 L

calculating the new concentrated value of NH_3 = \frac{0.10\times 0.040}{0.120}= 0.33 \ M

calculating the new concentrated value of NH_4Cl= \frac{0.050\times 0.10}{0.120}= 0.04166 \ Mcalculating the new concentrated value of H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M when 1 mol H_2So_4 produced 2 mols H^{+} so, 0.0125 in H_2So_4produced:

=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}

create the ICE table:    

NH_3    \ \ \ \ \ \ \ \     + H^{+}  \ \ \ \ \ \ \longrightarrow NH_4^{+}                    

I (m)       0.033(m)            0.025                       0.04166

C            -0.025                 -0.025                       + 0.025  

E            8.3\times 10^{-3}     0                    0.0667

now calculating pH:

when ph= 8.83:

P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769

5 0
3 years ago
If 5.12 liters of a 2.75 M phosphoric acid is neutralized by magnesium hydroxide solution of 4.00 M.
Tresset [83]

Answer:

5.28 L

Explanation:

Step 1:

Data obtained from the question.

Volume of acid (Va) = 5.12L

Molarity of acid (Ma) = 2.75M

Molarity of base (Mb) = 4M

Volume of base (Vb) =.?

Step 2:

The balanced equation for the reaction

2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O

From the balanced equation above,

Mole ratio of the acid (nA) = 2

Mole ratio of the base (nB) = 3

Step 3:

Determination of the volume of the base.

This is illustrated below:

MaVa/MbVb = nA/nB

2.75 x 5.12 / 4 x Vb = 2/3

Cross multiply

4 x 2 x Vb = 2.75 x 5.12 x 3

Divide both side by 4 x 2

Vb = (2.75 x 5.12 x 3)/(4 x 2)

Vb = 5.28 L

Therefore, the volume of the base is 5.28 L

8 0
3 years ago
Read 2 more answers
A flask with a volume of 125.0mL contains air with a density of 1.269 g/L. What is the mass of the air contained in the flask?
Aliun [14]
<span>In order to solve this problem you must first make sure all your numbers are in like terms. From the density value you can see that it is grams per liter. The first conversion you must do in convert the 125.0 mL value to Liters which you would do by dividing by 1000 because 1 liter is equal to 1000 mL. 125.0 divided by 1000 is 0.125 Liter. Now you will use the density equation to solve. The density equation is density is equal to mass divided by volume. Plug in your known numbers for density and volume. Then solve for mass. So Density (1.269 g/l is equal to mass divided by volume (.125 Liter) You must rearrange the equation to multiple density by volume which is 1.269 times 0.125 which will give you 0.1586. Because the Liters cancel each other out, the answer's unit will be grams. Your final answer is 0.1586 grams.</span>
4 0
2 years ago
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