Answer:
1) <em>The correct answer is A. Collision</em>
2) A hot solvent helps a solid dissolve faster because an increase in <u><em>kinetic energy</em></u> that also increases the rate of collisions
Explanation:
When a solute is added into a solvent and stirred, the solute particles get distributed to all parts of the solvent as a result of stirring.
More collisions occur between the solute and the solvent due to stirring. This increases the rate of dissolving.
<em>When a solvent is heated, then the kinetic energy would increase and the atoms will collide with a much greater force. As a result, ore solute will be able to dissolve in the solvent. </em>
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
Answer:
B. two rounds of cell division
Answer : The temperature of liquid is, 369.9 K
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of liquid at 373 K = 681 torr
= vapor pressure of liquid at normal boiling point = 760 torr
= temperature of liquid = ?
= normal boiling point of liquid = 373 K
= heat of vaporization = 40.7 kJ/mole = 40700 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the temperature of liquid is, 369.9 K
Answer:
Explanation:
In general, an increase in pressure (decrease in volume) favors the net reaction that decreases the total number of moles of gases, and a decrease in pressure (increase in volume) favors the net reaction that increases the total number of moles of gases.
Δn= b - a
Δn= moles of gaseous products - moles of gaseous reactants
Therefore, <u>after the increase in volume</u>:
- If Δn= −1 ⇒ there are more moles of gaseous reactants than gaseous products. The equilibrium will be shifted towards the products, that is, from left to right, and K>Q.
- If Δn= 0 ⇒ there is the same amount of gaseous moles, both in products and reactants. The system is at equilibrium and K=Q.
- Δn= +1 ⇒ there are more moles of gaseous products than gaseous reactants. The equilibrium will be shifted towards the reactants, that is, from right to left, and K<Q.
