Answer:
The ΔHrxn for the above equation = 179 kJ/mol
Explanation:
The reaction bond enthalpies are for the reactant;
3 × N-H = 3 × 390 = 1,170 kJ/mol
2 × O=O = 2 × 502 = 1004 kJ/mol
The reaction bond enthalpies are for the product;
3 × N-O = 3 × 201 = 603 kJ/mol
3 × O-H = 3 × 464 = 1,392 kJ/mol
The ΔHrxn for the above equation is therefore;
ΔHrxn = 1,170 + 1,004 - (603 + 1,392) = 179 kJ/mol
Answer:
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
Explanation:
Step 1: Data given
Volume = 500 mL = 0.500 L
The concentration sodium sulfate = 2.104 M
Step 2: The equation
Na2SO4 → 2Na+ + SO4^2-
For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)
Step 3: Calculate the concentration of the ions
[Na+] = 2*2.104 M = 4.208 M
[SO4^2-] = 1*2.104 M = 2.104 M
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
Oxygen,hydrogen, and carbon
The size, length, and height of the figure are the main factors which determine the shadow of a organism.
