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Damm [24]
2 years ago
13

The octet rule is the basis for the stability of the atom. According to this rule, the outermost shell of the atom must contain

__________ electrons.
Chemistry
1 answer:
liubo4ka [24]2 years ago
6 0

Answer:8

Explanation:

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Mass has to do with how many particles
Ganezh [65]

The law of definite proportions agrees with Dalton atomic theory.

What is Dalton atomic theory?

It state that all matters is made of very tiny particles called atom. atoms are individual particles which can not be created or be destroyed in a chemical reactions. Atoms of given elements are identical in mass and chemical properties. Atoms of

different elements have different masses and chemical properties.

The law of definite proportions also known as proust's law ,state that a chemical compound contain the same proportion of elements by mass.this law is one of the stoichiometry .

Thus ,

This is the reason why it is agrees with dalton atomic theory.

To know more about Dalton atomic theory click-

brainly.com/question/13157325

#SPJ1

6 0
1 year ago
Which power source uses the earth’s internal heat?
Nataly_w [17]

Answer:

Its geothermal energy.

Explanation:

hope this helps

5 0
3 years ago
(C6H6) can be biodegraded by microorganism. if 30 mg of benzene is present, what amount of oxygen required for biodegradation, n
quester [9]

Answer:

36.92 mg of oxygen required for bio-degradation.

Explanation:

5C_6H_6+15O_2\rightarrow 12CO_2+15H_2O

Mass of benzene = 30 mg = 0.03 g (1000 mg = 1 g )

Moles benzene =\frac{0.03 g}{78 g/mol}=0.0003846 mol

According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.

Then 0.0003846 mol of benzene will react with:

\frac{15}{5}\times 0.0003846 mol=0.0011538 mol of oxygen gas

Mass of 0.0011538 moles of oxygen gas:

0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg

36.92 mg of oxygen required for bio-degradation.

4 0
3 years ago
Which method do you consider most appropriate for determining half-life?
Masja [62]
It's the person that you in love with.<span />
3 0
3 years ago
5. Durante un estudio de la velocidad de la reacción A2(g) + 3B2(g)  2 AB3(g), se observa que en un recipiente cerrado que cont
weqwewe [10]

Answer:

a) Speed of the reaction = 0.002083 mol/L.s

b) The rate of disappearance of A₂ during this period of time = 0.002083 mol/L.s

c) The rate of appearance of AB₃ = 0.004167 mol/L.s

Explanation:

English Translation

During a study of the reaction rate

A₂ (g) + 3B₂ (g) → 2 AB₃ (g),

it is observed that in a closed container containing a certain amount of A₂ and 0.75 mol / L of B₂, the concentration B₂ decreases to 0.5 mol / L in 40 seconds.

a) What is the speed of the reaction?

b) What is the rate of disappearance of A₂ during this period of time?

c) What is the rate of appearance of AB₃?

Solution

The rate of a chemical reaction is defined as the time rate at which a reactant is used up or the rate at which a product is formed.

It is the rate of change of the concentration of a reactant (rate of decrease of the concentration of the reactant) or a product (rate of increase in the concentration of the product) with time.

Mathematically, for a balanced reaction

aA → bB

Rate = -(1/a)(ΔA/Δt) = (1/b)(ΔB/Δt)

The minus sign attached to the change of the reactant's concentration indicates that the reactant's concentration decreases.

And the coefficients of each reactant and product in the balanced reaction normalize the rate of reaction for each of them

So, for our given reaction,

A₂ (g) + 3B₂ (g) → 2 AB₃ (g)

Rate = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt) = (1/2)(ΔAB₃/Δt)

a) Speed of the reaction = Rate of the reaction

But we are given information on the change of concentration of B₂

Change in concentration of B₂ = ΔB₂ = 0.50 - 0.75 = -0.25 mol/L

Change in time = Δt = 40 - 0 = 40 s

(ΔB₂/Δt) = (-0.25/40) = -0.00625 mol/L.s

Rate of the reaction = -(1/3)(ΔB₂/Δt) = (-1/3) × (-0.00625) = 0.002083 mol/L.s

b) The rate of disappearance of A₂ during this period of time

Recall

Rate = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt)

-(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt)

Rate of disappearance of A₂ = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt) = (-1/3) × (-0.00625) = 0.002083 mol/L.s

c) The rate of appearance of AB₃

Recall

Rate = -(1/3)(ΔB₂/Δt) = (1/2)(ΔAB₃/Δt)

(1/2)(ΔAB₃/Δt) = -(1/3)(ΔB₂/Δt)

(ΔAB₃/Δt) = -(2/3)(ΔB₂/Δt)

rate of appearance of AB₃ = (ΔAB₃/Δt) = -(2/3)(ΔB₂/Δt) = (-2/3) × (-0.00625) = 0.004167 mol/L.s

Hope this Helps!!!

3 0
3 years ago
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