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3 years ago

What is the octet rule and how is it used in covalent bonding?

1 answer:

4 0

The octet rule is the fundamental explanation to give the reasons as for why atoms would undergo complex molecular and ionic arrangements, by sharing or transferring valence or bonding electrons between atoms in a compound.

It states that atoms would need to have the maximum number of electrons in their valence shell to be inactive and stable. Thus atoms would tend to bond to other atoms to achieve this Lower energy and stability.

They do this by sharing between electrons between atoms as in covalent bonds.

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What happens when grapes get fermented

ValentinkaMS [17]

When grapes become fermented, they turn into wine.

6 0

3 years ago

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

If the rate of decrease for the partial pressure of N2H4N2H4 in a closed reaction vessel is 70 torr/htorr/h , what is the rate o

german

Answer:

$r_{NH_3}=140torr/h$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$N_2H_4 (g) + H_2 (g) \rightarrow 2 NH_3$

Thus, in terms of pressures, the rate becomes:

$-r_{N_2H_4}=\frac{1}{2} r_{NH_3}$

Thus, the rate of change for the partial pressure of ammonia turns out:

$r_{NH_3}=2*(-r_{N_2H_4})\\r_{NH_3}=2*[-(-70torr/h)]\\r_{NH_3}=140torr/h$

The rate of decrease of partial pressure of urea is taken negative as it is a reactant whereas ammonia a product which has 2 as its stoichiometric coefficient.

Best regards.

7 0

3 years ago

Nic bought 12 pens for 2$ each and 12pads of paper for 3$ each. Write two expressions for how nic could calculate his total.Writ

DENIUS [597]

12 pens for writing chemical formulas cost
12*2=24 dollars
12 pads cost
12*3=36 dollars
1)
12*2=24
12*3=36
2)
12*2+12*3=60 dollars
3) 12*(3+5)=60 dollars

7 0

3 years ago

What happens in a chemical reaction with the individual elements?

Gemiola [76]

Chemical reactions involve electrons and where they move (if they move).
Metal atoms (or elements) usually give away their outer shell electrons in reactions to become stabilised (creating positive ions). Non-metal atoms (or elements) usually gain electrons in order to become stable (negative ions). Some reactions involve covalent bonds, metallic bonds, ionic bonds etc., but at the base (no pun intended :D) of most chemical reactions it's just a matter of electrons moving around between atoms.

Hope it helps!

3 0

4 years ago