The correct option is (a) H₂S : SO₂ = 2 : 2 and O₂ : H₂O = 3 : 2
<h3>What is molar Ratio ?</h3>
Molar ratio also known as stoichiometry is the ratio in which the reactants and products are either formed or reacted in the given equation
the balanced equation is as follows ;
2H₂S + 3O₂ --> 2SO₂ + 2H₂O
molar ratio can be determined by the coefficients of the compounds in the balanced reaction.
coefficient is the number in front of the chemical compound
coefficients for the compounds in this reaction are as follows ;
- H₂S - 2
- O₂ - 3
- SO₂ - 2
- H₂O - 2
therefore, correct option is (a) H₂S:SO₂ = 2:2 and O₂:H₂O = 3:2
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Answer:
0.482 ×10²³ molecules
Explanation:
Given data:
Volume of gas = 2.5 L
Temperature of gas = 50°C (50+273 = 323 k)
Pressure of gas = 650 mmHg (650/760 =0.86 atm)
Molecules of N₂= ?
Solution:
PV= nRT
n = PV/RT
n = 0.86 atm × 2.5 L /0.0821 atm. mol⁻¹. k⁻¹. L × 323 k
n = 2.15 atm. L /26.52 atm. mol⁻¹.L
n = 0.08 mol
Number of moles of N₂ are 0.08 mol.
Number of molecules:
one mole = 6.022 ×10²³ molecules
0.08×6.022 ×10²³ = 0.482 ×10²³ molecules
Answer:
4.81 moles
Explanation:
The total pressure of the gas = Pressure at which gauge reads zero + pressure read by it.
Pressure at which gauge reads zero = 14.7 psi
Pressure read by the gauge = 988 psi
Total pressure = 14.7 + 988 psi = 1002.7 psi
Also, P (psi) = P (atm) / 14.696
Pressure = 1002.7 / 14.696 = 68.2297 atm
Temperature = 25 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25 + 273.15) K = 298.15 K
Volume = 1.50 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K
⇒n = 4.81 moles
Answer:
composting scraps
recycling is the action or process of converting waste into reusable material.
Answer:
Explanation:
For the reaction
C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O
We can calculate the standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:
ΔHºc = 2ΔHºf CO2 (g) + 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )
( we were not give the water state but we know we are at standard conditions so it is in its liquid state )
The ΔHºfs can be found in appropiate reference or texts.
ΔHºc = 2ΔHºf CO2 (g)+ 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )
= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [( -276.2 + 0 ) ] kJ
ΔHºc = -1368.33 kJ
