Notice that in the equation given, there is 1 mol of Mg3N2
and 6 mol of H2O. Consequently, the question is requesting, if we took 2.5 mol
of Mg3N2, how many mol of H2O would be essential to complete a similar reaction
-- in quintessence, multiplying everything by 2.5:
the solution would be: 6 * 2.5 = 15 mol
Answer: Parts per million (ppm)
Explanation:
Consider the units milligram per milliliter. This gives us one part of the solute per one million parts of solvent. That is 10^ -3/10^-3= 10^-6. This unit is commonly used in analytical chemistry to show very small concentration of analyte. A similar unit is parts per billion(ppb)
Answer:
ΔHrxn = -635.14kJ/mol
Explanation:
We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:
(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol
(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole
(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole
Reaction (1) - (3) produce:
Ca(s) + H2O(l) → H2(g) + CaO(s)
ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol
Now this reaction + 1/2(2):
Ca(s) + ½ O2(g) → CaO(s)
ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)
<h3>ΔHrxn = -635.14kJ/mol</h3>
Answer:
78.9% yield
Explanation:
% Yield = Actual Lab Yield/Theoretical Yield x 100%
= (204.6g)/(256.5g)·100% = 79.8% Yield
It is D. because the student is using a <span> thermometer and a rain gauge </span>
