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Veronika [31]
3 years ago
5

Which is a natural​ heat-producing process that may give rise to a​ fire?

Chemistry
1 answer:
tigry1 [53]3 years ago
5 0
B is correct.  Please Mark Brainliest!!!
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What is the total number of outer (valence) electrons in carbon dioxide, co2? answer?
Anuta_ua [19.1K]
Carbon dioxide has a total of 16 valence electrons.

1. To determine the number of valence electrons of carbon dioxide (CO2), first determine the number of valence electrons of each of the elements in the molecule. 

a. We have 1 carbon (C) molecule, and 2 oxygen (O) molecules. 

b. The carbon molecule has 4 valence electrons and each oxygen molecule has 6 oxygen molecules. 

2. Add up the valence electrons of each of the elements

   4            +          (2 x 6)      = 16
(from C)           (2 oxygen molecules, with 6 valence electrons each)

Thus, CO2 has a total of 16 valence electrons. 

The number of valence electrons can be more clearly seen from the Lewis structure of the CO2 in the figure below (Source: http://chemistry.tutorvista.com/inorganic-chemistry/bonding-electrons.html). The the dots surrounding the letters represent the valence electrons. 

8 0
3 years ago
Which chemical equation demonstrates the law of conservation of mass!
zlopas [31]

Answer:

F. 2NO + 02 —> 2NO

H. 4NH3 + 502 —> 4NO + 6H20

Explanation:

The law of conservation of mass states that matter can neither be created nor destroyed during a chemical reaction but can be convert from one form to another.

2NO + 02 —> 2NO

From the above, the total number of N on the left balance the total number on the right i.e 2 atoms of N on both side of the equation.

The total number of O on the left balance the total number on the right i.e 2 atoms of O on both side of the equation. This is certified by the law of conservation of mass.

4NH3 + 502 —> 4NO + 6H20

From the above, the total number of N on the left balance the total number on the right i.e 4 atoms of N on both side of the equation.

The total number of O on the left balance the total number on the right i.e 10 atoms of O on both side of the equation.

The total number of H on the left balance the total number on the right i.e 12 atoms of O on both side of the equation.

This is certified by the law of conservation of mass.

The rest equation did not conform to the law of conservation of mass as the atoms on the left side did not balance those on the right side

5 0
3 years ago
A chemist fills a reaction vessel with 0.750 M lead (II) (Pb2+) aqueous solution, 0.232 M bromide (Br) aqueous solution, and 0.9
Ronch [10]

Answer:

The free energy = -20.46 KJ

Explanation:

given Data:

Pb²⁺ = 0.750 M

Br⁻ = 0.232 M

R = 8.314 Jk⁻¹mol⁻¹

T = 298K

The Gibb's free energy is calculated using the formula;

ΔG = ΔG° + RTlnQ -------------------------1

Where;

ΔG° = standard Gibb's freeenergy

R = Gas constant

Q = reaction quotient

T = temperature

The chemical reaction is given as;

Pb²⁺(aq) + 2Br⁻(aq) ⇄PbBr₂(s)

The ΔG°f are given as:

ΔG°f (PbBr₂)  = -260.75 kj.mol⁻¹

ΔG°f (Pb²⁺)   = -24.4 kj.mol⁻¹

ΔG°f (2Br⁻)    = -103.97 kj.mol⁻¹

Calculating the standard gibb's free energy using the formula;

ΔG° = ξnpΔG°(product) - ξnrΔG°(reactant)

Substituting, we have;

ΔG° =[1mol*ΔG°f (PbBr₂)] - [1 mol *ΔG°f (Pb²⁺) +2mol *ΔG°f (2Br⁻)]

ΔG° =(1 *-260.75 kj.mol⁻¹) - (1* -24.4 kj.mol⁻¹) +(2*-103.97 kj.mol⁻¹)

      = -260.75 + 232.34

     = -28.41 kj

Calculating the reaction quotient Q using the formula;

Q = 1/[Pb²⁺ *(Br⁻)²]

   = 1/(0.750 * 0.232²)

  = 24.77

Substituting all the calculated values into equation 1, we have

ΔG = ΔG° + RTlnQ

ΔG = -28.41 + (8.414*10⁻³ * 298 * In 24.77)

     = -28.41 +7.95

    = -20. 46 kJ

Therefore, the free energy of reaction = -20.46 kJ

8 0
3 years ago
Net ionic of ammonium sulfide added to iron (ll) chloride
meriva

Answer:

Fe²⁺(aq) + S²⁻(aq )⟶ FeS(s)  

Step-by-step explanation:

Molecular Equation:

(NH₄)₂S(aq) + FeCl₂(aq) ⟶ 2NH₄Cl(aq) + FeS(s)

Ionic equation :

2NH₄⁺(aq) + S²⁻(aq) + Fe²⁺(aq) + 2Cl⁻(aq) ⟶ 2NH₄⁺(aq) + 2Cl⁻(aq) + FeS(s)

Net ionic equation :

Cancel all ions that appear on both sides of the reaction arrow (underlined).

<u>2NH₄⁺(aq)</u> + S²⁻(aq) + Fe²⁺(aq) + <u>2Cl⁻(aq)</u> ⟶ <u>2NH₄⁺(aq) </u>+ 2<u>Cl⁻(aq) </u>+ FeS(s)

Fe²⁺(aq) + S²⁻(aq )⟶ FeS(s)

4 0
3 years ago
I need help with this chem study guide
Cloud [144]
Can you please provide a clearer image?
Thanks
7 0
3 years ago
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