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diamong [38]
3 years ago
13

H2 + O2 -> 2H2O balanced or unbalanced pls help

Chemistry
1 answer:
Sidana [21]3 years ago
7 0

Answer:

Unbalanced

Explanation:

The reaction equation is given as:

             H₂  +  O₂  →   2H₂O

The given reaction flouts the law of conservation of mass and it is not balanced. To know a reaction that is balanced, the number of atoms on both sides of the expression must be the same;

                                             Number of moles

                         Reactants                                             Products

Hydrogen                2                                                            4

Oxygen                   2                                                            2

We see that the hydrogen atoms are not balanced.

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Which unit can be used to express the concentration of a solution?
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Answer: (3) ppm

Explanation: Concentration is defined as the amount of solute dissolved in a known amount of the solvent or  a solution.

Parts per million (ppm) is used to express the concentration when a very small quantity of solute is present in a large quantity of the solution. It is defined as the mass of solute present in one milion (10^6) parts by mass of the solution.

ppm=\frac{\text {mass of Solute}}{\text {Mass of solution}}\times 10^6

L/s is used to express flow rate.

J/g is used to express energy per unit mass.

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6 0
3 years ago
Read 2 more answers
Calculate the values of ΔU, ΔH, and ΔS for the following process:
ladessa [460]

Answer:

ΔU = 45.814 KJ

ΔH = 46.4375 KJ

ΔS = 18.76 J/K

Explanation:

            H2O(l)        →          H2O(l)                →              H2O(steam)

   298.15K, 1atm   ΔHp     373.15K,1atm       ΔHv         373.15K,1 atm

∴ ΔHp = Qp = nCpΔT

∴ n H2O = 1 mol

∴ Cp,n = 75.3 J/mol.K

∴ ΔT = 373.25 - 298.15 = 75 K

⇒ Qp = (1 mol)*(75.3 J/mol.K)*(75K) = 5647.5 J

⇒ ΔHp = 5647.5 J = 5.6475 KJ

⇒ ΔH = ΔHp + ΔHv

∴ ΔHv = 40.79 KJ/mol * 1 mol = 40.79 KJ  

⇒ ΔH = 5.6475 KJ + 40.79 KJ = 46.4375 KJ

ideal gas:

∴ ΔH = ΔU + PΔV

∴ V1 = nRT1/P1 = ((1)*(0.082)*(298.15))/1 = 24.45 L

∴ V2 = nRT2/P2 = ((1)*(0.082)*(373.15))/ 1 = 30.59 L

⇒ ΔV = V2 - V1 = 6.15 L * (m³/1000L) = 6.15 E-3 m³

∴ P = 1 atm * (Pa/ 9.86923 E-6 atm) = 101325.027 Pa

⇒ ΔU = ΔH - PΔV = 46.4375 KJ - ((101325.027 Pa*6.15 E-3m³)*(KJ/1000J))

⇒ ΔU = 46.4375 KJ - 0.623 KJ

⇒ ΔU = 45.814 KJ

∴ ΔS = Cv,n Ln (T2/T1) + nR Ln (V2/V1)

⇒ ΔS = (75.3) Ln(373.15/298.15) + (1)*(8.314) Ln (30.59/24.45)

⇒ ΔS = 16.896 J/K + 1.863 J/K

⇒ ΔS = 18.76 J/K

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