Answer: The enthalpy change is 34.3 kJ
Explanation:
The conversions involved in this process are :
Now we have to calculate the enthalpy change.
![\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= enthalpy change = ?
m = mass of water = 72.0 g
= specific heat of ice = 
= specific heat of liquid water = 
n = number of moles of water = 
= enthalpy change for fusion = 6010 J/mole
Now put all the given values in the above expression, we get
![\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B72.0g%5Ctimes%202.09J%2Fg%5E0C%5Ctimes%20%280-%28-18%29%5E0C%5D%2B4.00mole%5Ctimes%206010J%2Fmole%2B%5B72.0g%5Ctimes%204.184J%2Fg%5E%29C%5Ctimes%20%2825-0%29%5E0C%5D)
(1 KJ = 1000 J)
Therefore, the enthalpy change is 34.3 kJ
Question:
a. Diffusion
b. Facilitated diffusion
c. Both
d. Neither
1. movement to area of lower concentration
2. movement across a membrane
3. steroid transport into cell
4. requires energy
5. movement assisted by proteins
6. glucose transport into cell
Answer:
The sorting is as follows
a. (1)
b. (5 and 6)
c. (1 and 2)
d. (4)
Explanation:
Diffusion is the movement of particles across a membrane from a high concentration region to one with a lower concentration of the diffusing substance
Here we have the correct sorting as follows
a. Diffusion
3. steroid transport into cell
b. Facilitated diffusion
5. movement assisted by proteins
6. glucose transport into cell
c. Both
1. movement to area of lower concentration
2. movement across a membrane
d. Neither
4. requires energy
The interaction between the substances and stomata is a chemical reaction, light energy strikes at the chlorophyll molecules, whose electrons gets excited to a state of higher energy, and they come back to their state of lower energy by emission of energy, which is accepted by a chain of acceptors, and energy is generated.
It's a hazard symbol for irritant.
Answer:
Q14: 17,140 g = 17.14 kg.
Q16: 504 J.
Explanation:
<u><em>Q14:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = 3600 x 10³ J).
m is the mass of the ice (m = ??? g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 0.0°C = 100.0°C).
∵ Q = m.c.ΔT
∴ (3600 x 10³ J) = m.(2.1 J/g.°C).(100.0°C)
∴ m = (3600 x 10³ J)/(2.1 J/g.°C).(100.0°C) = 17,140 g = 17.14 kg.
<u><em>Q16:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = ??? J).
m is the mass of the ice (m = 12.0 g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 0.0°C - (-20.0°C) = 20.0°C).
∴ Q = m.c.ΔT = (12.0 g)(2.1 J/g.°C)(20.0°C) = 504 J.
