The question is incomplete. The complete question is :
Acetylene 
gas is often used in welding torches because of the very high heat produced when it reacts with oxygen 
gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.5 mol of carbon dioxide. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Answer: 1.9 moles of oxygen are needed to produce 1.5 moles of carbon dioxide.
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
The balanced reaction for combustion of acetylene is:
According to stoichiometry;
4 moles of carbon dioxide are produced by = 5 moles of oxygen
Thus 1.5 moles of carbon dioxide are produced by =
moles of oxygen
Thus 1.9 moles of oxygen are needed to produce 1.5 moles of carbon dioxide.
Answer:
ΔH°C2H2Cl4(l) = -333,36 kJ/mol
ΔH°r₂ = -35,14 kJ/mol
Explanation:
The ΔH°r of the first reaction is:
ΔH°r = -385,76 kJ/mol = (ΔH°C2H2Cl4(l) + ΔH°H2(g)) - (ΔH°C2H4(g) + 2ΔHCl2 (g))
ΔH°H2(g) = 0 kJ/mol
ΔH°C2H4(g) = 52,4 kJ/mol
Δ°HCl2 (g) = 0 kJ/mol
Replacing:
ΔH°C2H2Cl4(l) = -385,76 kJ/mol + 52,4 kJ/mol = <em>-333,36 kJ/mol</em>
The standard heat of the second reaction is:
ΔH°r₂ = ΔH°C2HCl3(l) + ΔH°HCl(g) - ΔH°C2H2Cl4(l)
Where:
ΔH°C2HCl3(l) = -276,2 kJ/mol; ΔH°HCl(g) = -92,3 kJ/mol; ΔH°C2H2Cl4(l) = -333,36 kJ/mol
Replacing:
ΔH°r₂ = -276,2 kJ/mol -92,3 kJ/mol + 333,36 kJ/mol
<em>ΔH°r₂ = -35,14 kJ/mol</em>
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I hope it helps!
Answer:
Q = -811440 J
Explanation:
Given data:
Mass of oil = 2.76 Kg (2.76× 1000 = 2760 g)
Initial temperature = 191 °C
Final temperature = 23°C
Specific heat capacity of oil = 1.75 J/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 23°C - 191 °C
ΔT = -168°C
Q = 2760 g ×1.75 J/g.°C ×-168°C
Q = -811440 J
Negative sign show heat is released.
The quality of being easily dissolved in liquid.
Hope this help and Happy holidays
Answer:
552 g of LiNO₃
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 2 L
Molarity of LiNO₃ = 4 M
Mass of LiNO₃ =?
Next, we shall determine the number of mole of LiNO₃ in the solution. This can be obtained as follow:
Volume of solution = 2 L
Molarity of LiNO₃ = 4 M
Mole of LiNO₃ =?
Molarity = mole /Volume
4 = mole of LiNO₃ / 2
Cross multiply
Mole of LiNO₃ = 4 × 2
Mole of LiNO₃ = 8 moles
Finally, we shall determine the mass of of LiNO₃ needed to prepare the solution. This is can be obtained as follow:
Mole of LiNO₃ = 8 moles
Molar mass of LiNO₃ = 7 + 14 + (16×3)
= 7 + 14 + 48
= 69 g/mol
Mass of LiNO₃ =?
Mole = mass /Molar mass
8 = Molar mass of LiNO₃ /69
Cross multiply
Molar mass of LiNO₃ = 8 × 69
Molar mass of LiNO₃ = 552 g
Thus, 552 g of LiNO₃ is needed to prepare the solution.
