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Ahat [919]
3 years ago
14

What makes the desert unique from other biomes.

Chemistry
1 answer:
Vlada [557]3 years ago
5 0

Explanation:

Desert biomes are the driest of all the biomes. In fact, the most important characteristic of a desert is that it receives very little rainfall. Most deserts receive less than 300 mm a year compared to rainforests, which receive over 2,000 mm.

NASA

plants in it can grow in harsh conditions and store moisture within the plants itself

sciencing

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I need help FAST ASAP
kiruha [24]
In this item, we are simply to find the ions that may bond and are able to form a formula unit. We are also instructed to give out their name. There are numerous possible combinations of ions to form a compound. Some answers are given in the list below.

1.  Na⁺     ,    Cl⁻    , NaCl   ---> sodium chloride (this is most commonly known as table salt)

2. C⁴⁺       , O²⁻     , CO₂  ---> carbon dioxide

3. Al³+     , Cl⁻       , AlCl₃   ----> aluminum chloride

4. Ca²⁺     , Cl⁻     , CaCl₂    ---> calcium chloride

5. Li⁺        , Br⁻      , LiBr       ---> lithium bromide

6. Mg³⁺     , O²⁻      , Mg₂O₃   ----> magnesium oxide

7. K⁺        , I⁻          , KI   ---> potassium iodide

8. H⁺        , Cl⁻        , HCl  --> hydrogen chloride

9. H⁺        , Br⁻         , HBr ----> hydrogen bromide

10. Na⁺    , Br⁻         , NaBr   ---> sodium bromide
6 0
3 years ago
Scientists have observed that hurricanes are strongest over water, and that hurricanes lose momentum and become weak when they r
asambeis [7]

Answer:

Explanation:                                                                                        

4 0
2 years ago
What is the ph of a 0.25 m solution of c6h5nh2 given that its kb is 1.8 x 10-6?
IrinaK [193]

The pH a 0.25 m solution of C₆H₅NH₂ is equal to 3.13.

<h3>How do we calculate pH of weak base?</h3>

pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:

pH = pKb + log([HB⁺]/[B])

pKb = -log(1.8×10⁻⁶) = 5.7

Chemical reaction for C₆H₅NH₂ is:

                          C₆H₅NH₂ + H₂O → C₆H₅NH₃⁺ + OH⁻

Initial:                     0.25                           0            0

Change:                    -x                             x             x

Equilibrium:        0.25-x                           x             x

Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]

Kb = x² / 0.25 - x

x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:

1.8×10⁻⁶ = x² / 0.25

x² = (1.8×10⁻⁶)(0.25)

x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]

On putting all these values on the above equation of pH, we get

pH = 5.7 + log(0.67×10⁻³/0.25)

pH = 3.13

Hence pH of the solution is 3.13.

To know more about Henderson Hasselbalch equation, visit the below link:
brainly.com/question/13651361

#SPJ4

5 0
2 years ago
The electrolyte of the following is:
ozzi
An electrolyte is a term used to describe a compound that can dissociate into ions as it is nothing but an ionic compound, a salt made up of a positively charged cation and negatively charged anion.

Here the correct answer is D. Since there are no hydrocarbons or any other organic compound, that do not possess partial let alone full charges, all of them can dissociate in solution to give their ions.

This allows for the solution to be able to conduct electricity.
4 0
2 years ago
Which of the following acids should be used to prepare a buffer with a pH of 4.5?A. HOC6H4OCOOH, Ka = 1.0 x 10^-3B. C6H4(COOH)2,
ioda

Answer:

C. CH3COOH, Ka = 1.8 E-5

Explanation:

analyzing the pKa of the given acids:

∴ pKa = - Log Ka

A. pKa = - Log (1.0 E-3 ) = 3

B. pKa = - Log (2.9 E-4) = 3.54

C. pKa = - Log (1.8 E-5) = 4.745

D. pKa = - Log (4.0 E-6) = 5.397

E. pKa = - Log (2.3 E-9) = 8.638

We choose the (C) acid since its pKa close to the expected pH.

⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):

  • pH = pKa + Log ([CH3COO-]/[CH3COOH])

∴ pH = 4.5

∴ pKa = 4.745

⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])

⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])

⇒ 0.5692 = [CH3COO-]/[CH3COOH]

∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]

⇒ 1.8 E-5 = [H3O+](0.5692)

⇒ [H3O+] = 3.1623 E-5 M

⇒ pH = - Log ( 3.1623 E-5 ) = 4.5

6 0
2 years ago
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