The answer is D: 400N
your welcome
Answer:
204g of NH3
Explanation:
The balanced equation for the reaction is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the number of mole NH3 produced by reacting 6moles of N2. This is illustrated below:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 6 moles of N2 will react to produce = 6 x 2 = 12 moles of NH3.
Finally, we shall convert 12 moles of NH3 to grams. This is illustrated below:
Number of mole of NH3 = 12 moles.
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Mass of NH3 =..?
Mass = mole x molar mass
Mass of NH3 = 12 x 17
Mass of NH3 = 204g.
Therefore, 204g of NH3 will be produced from the reaction.
Answer:
ELEMENTS
COMPOUNDS
Elements are made up of one kind of atoms.
Compounds are made up of two or more kinds of atoms.
Elements cannot be broken down into simpler substances by any physical or chemical method.
Compounds can be broken down into simpler substances by chemical methods.
Elements have their own set of properties.
Properties of a compound differ from those of their elements.
Examples: Hydrogen, Oxygen
Examples: Water, Sodium chloride
<span>Answer:
Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either.
Start with your 2 half reactions:
I- --> IO3-
Cr2O72- --> 2 Cr3+
Balance O by adding H2O:
I- + 3 H2O --> IO3-
Cr2O72- --> 2 Cr3+ + 7H2O
Balance H by adding H+:
I- + 3 H2O --> IO3- + 6 H+
Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O
Balance charge by adding e-:
I- + 3 H2O --> IO3- + 6 H+ + 6 e-
Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O
Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give:
Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O</span>
Enthalpy of formation is calculated by subtracting the total enthalpy of formation of the reactants from those of the products. This is called the HESS' LAW.
ΔHrxn = ΔH(products) - ΔH(reactants)
Since the enthalpies are not listed in this item, from reliable sources, the obtained enthalpies of formation are written below.
ΔH(C2H5OH) = -276 kJ/mol
ΔH(O2) = 0 (because O2 is a pure substance)
ΔH(CO2) = -393.5 kJ/mol
ΔH(H2O) = -285.5 kJ/mol
Using the equation above,
ΔHrxn = (2)(-393.5 kJ/mol) + (3)(-285.5 kJ/mol) - (-276 kJ/mol)
ΔHrxn = -1367.5 kJ/mol
<em>Answer: -1367.5 kJ/mol</em>
