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9966 [12]
3 years ago
15

Which of the following is a balanced chemical equation?

Chemistry
2 answers:
ella [17]3 years ago
6 0
The balanced chemical equation is P4+3O2--> 2P2O3
SVEN [57.7K]3 years ago
6 0

Answer : The correct option is,

P_4+3O_2\rightarrow 2P_2O_3

Explanation :

Balanced chemical reaction : It is defined as the number of atoms of individual elements present on reactant side must be equal to the product side.

The given unbalanced chemical reaction is,

MgF_2+Li_2CO_3\rightarrow MgCO_3+LiF

CH_4+2O_2\rightarrow CO_2+H_2

2Al+6HCl\rightarrow 3H_2+AlCl_3

These chemical reactions are the unbalanced reactions because in this reaction, the number of atoms of individual elements present on reactant side are not equal to the product side.

Thus, the balanced chemical reaction will be,

P_4+3O_2\rightarrow 2P_2O_3

This chemical reaction is a balanced chemical reaction because in this reaction, the number of atoms of individual elements present on reactant side are equal to the product side.

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4 0
3 years ago
What is the compound name of FeBrO3?
AfilCa [17]
<h2>Answer:   Iron (I) Bromate</h2>

Explanation:

Fe⁺ - Iron (I)

BrO₃⁻ - Bromate

That makes the compount Iron (I) Bromate

3 0
3 years ago
Calculate the molarity of a solution that contains 0.37g CaCl2 in 340 milliliters of solution
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Answer:

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Explanation:

3 0
2 years ago
Calculate Ecell for the following electrochemical cell at 25 ºCPt (s) | H2 (g, 1.00 atm) | H+ (aq, 1.00 M) || Sn2+ (aq, 0.350 M)
katrin2010 [14]

<u>Answer:</u> The electrode potential of the cell is 0.093 V

<u>Explanation:</u>

The given chemical cell follows:

Pt(s)|H_2(g,1atm)|H^+(aq,1.00M)||Sn^{4+}(aq,0.020M)|Sn^{2+}(aq.,0.350M)|Pt(s)

<u>Oxidation half reaction:</u> H_2(g,1atm)\rightarrow 2H^{+}(aq,1.0M)+2e^-;E^o_{2H^{+}/H_2}=0.0V

<u>Reduction half reaction:</u> Sn^{4+}(aq,0.020M)+2e^-\rightarrow Sn^{2+(aq.,0.350M);E^o_{Sn^{4+}/Sn^{2+}}=0.13V

<u>Net cell reaction:</u>

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.13-(0.0)=0.13V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^2[Sn^{2+}]}{[Sn^{4+}]}

where,

E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = +0.13 V

n = number of electrons exchanged = 2

[H^{+}]=1.00M

[Sn^{2+}]=0.350M

[Sn^{4+}]=0.020M

Putting values in above equation, we get:

E_{cell}=0.13-\frac{0.059}{2}\times \log(\frac{(1.0)^2\times 0.350}{0.020})

E_{cell}=0.093V

Hence, the electrode potential of the cell is 0.093 V

3 0
4 years ago
What is the name of these 4 compound in isomers​
Lana71 [14]

Answer:

C9H19 C5H12 C6H17 C7H20

8 0
3 years ago
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