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4 years ago

Which stage is the middle stage of speciation? (15 Points)

Chemistry

1 answer:

goldenfox [79]4 years ago

7 0

The populations are isolated and thus do not interbreed.

Answer:B

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Ch3cooh (acetic acid) can form hydrogen bonds between its molecules. based on the lewis structure shown below, how many hydrogen

never [62]

Answer:
Number of Hydrogen Bond Acceptor atoms = 2

Number of Hydrogen Bond Donor atoms = 1

Explanation:
Hydrogen bond interactions are formed between those molecules which contains partial positive hydrogen atoms bonded covalently to most electronegative atoms like Oxygen, Nitrogen and Fluorine.
When hydrogen is attached to Oxygen, Nitrogen or Fluorine its electron density decreases and gets partial positive charge, this partial positive charged hydrogen atom then makes hydrogen bonding with the most electronegative element (partial negative) of neighbor molecule.
In Acetic acid there are two oxygen atoms hence there are two most electronegative elements therefore, two Hydrogen Bond Acceptor atom and each oxygen atom can accept two hydrogen bonds.
Also, it contains only one Hydrogen atom attached to oxygen atom so it has one Hydrogen Bond Donor atom.

4 0

3 years ago

Please hurry I’m stuck!!

MakcuM [25]

Answer:

Explanation:

C+O2=CO2

5 0

3 years ago

Read 2 more answers

Calculate the pH of a solution that contains 2.4 × 10 -5 M H 3O at 25°C. 9.38 11.60 4.62 4.17 2.40

lutik1710 [3]

Answer:

4.62

Explanation:

Using the equation -log[H3O+] = pH,

-log(2.4 x 10^-5) = 4.62

3 0

3 years ago

Read 2 more answers

1. A student needs to make 80mL of a 4.5 x 10^-3 M solution of H_3PO_4 (mm = 98) from solid. Describe how the student would do t

slamgirl [31]

Answer:

1. 35 mg of H₃PO₄

2. 27 mol AlF₃; 82 mol F⁻

3. 300 mL of stock solution.

Explanation:

1. Preparing a solution of known molar concentration

Data:

V = 80 mL

c = 4.5 × 10⁻³ mol·L⁻¹

Calculations:

(a) Moles of H₃PO₄

Molar concentration = moles of solute/litres of solution

c = n/V

n = Vc = 0.080L × (4.5 × 10⁻³ mol/1 L) = 3.60 × 10⁻⁴ mol

(b) Mass of H₃PO₄

moles = mass/molar mass

n = m/MM

m = n × MM = 3.60 × 10⁻⁴ mol × (98 g/1 mol) = 0.035 g = 35 mg

(c) Procedure

Dissolve 35 mg of solid H₃PO₄ in enough water to make 80 mL of solution,

2. Moles of solute.

Data:

V = 4900 mL

c = 5.6 mol·L⁻¹

Calculations:

Moles of AlF₃ = cV = 4.9 L AlF₃ × (5.6 mol AlF₃/1L AlF₃) = 27 mol AlF₃

Moles of F⁻ = 27 mol AlF₃ × (3 mol F⁻/1 mol AlF₃) = 82 mol F⁻.

3. Dilution calculation

Data:

V₁= 750 mL; c₁ = 0.80 mol·L⁻¹

V₂ = ? ; c₂ = 2.0 mol·L⁻¹

Calculation:

V₁c₁ = V₂c₂

V₂ = V₁ × c₁/c₂ = 750 mL × (0.80/2.0) = 300 mL

Procedure:

Measure out 300 mL of stock solution. Then add 500 mL of water.

3 0

4 years ago

Nitrobenzene contains 5 hydrogen atoms 6 carbon atoms 2 oxygen atoms and 1 nitrogen atom. what is the chemical formula for nitro

tamaranim1 [39]

To develop the chemical formula of a substance or compound in this regard, one only needs to write the chemical symbol for each of the element present and write as subscript to each symbol the number of atoms of that element. For nitrobenzene, as described above, the chemical formula should be,
C₆H₅O₂N

3 0

3 years ago