An object with a mass of 3.5 g raises the level of water in a graduated cylinder from 25.1 mL to 31.7 mL. What is the density of
1 answer:
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Answer:
Carbon 3 is double bonded to an oxygen and attached to carbon 2 and carbon 4. :
Answer: Carbonyl group ( Ketone or aldehyde)
Carbon 17 is attached to an oxygen, which is attached to a hydrogen. :
Answer: Carboxyl group (Carboxylic acid)
A central carbon is attached to an amine, two hydrogens, and a carbon that is double bonded to an oxygen and single bonded to an oxygen attached to a hydrogen. :
Answer: Amide group
An amide group contains both amine and carboxyl
The answer is 27.92g . working is shown in the picture above.
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K =
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M