Answer:
Ke = 34570.707
Explanation:
- H2(g) + Br2(g) → 2 HBr(g)
equilibrium constant (Ke):
⇒ Ke = [HBr]² / [Br2] [H2]
∴ [HBr] = (37.0 mol) / (2 L) = 18.5 mol/L
∴ [Br2] = (0.110 mol) / (2 L) = 0.055 mol/L
∴ [H2] = (0.360 mol) / (2 L) = 0.18 mol/L
⇒ Ke = (18.5 mol/L)² / (0.055 mol/L)(0.18 mol/L)
⇒ Ke = 34570.707
The compound that would have the highest osmotic pressure when dissolved in water is 
.
So, option D is correct one.
The dissociation of one molecule of gives the maximum number of ions when dissolved in water ( 4 ions ) . Osmotic pressure is a colligative property and depends upon number of solute particles present in the solution . The solution having maximum number of solute particles will have maximum number of the osmotic pressure .
All other given molecules gives less number of number of ions when dissolved in water as compare to of .
To learn more about osmotic pressure
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Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
Answer: C) middle 50 percent of the data
The interquartile range (IQR) spans from the first quartile Q1 to the third quartile Q3.
25% of the data is below Q1 and 75% of the data is below Q3. The gap between the two endpoints consists of 75-25 = 50 percent of the data, or half of the data.
