Answer:
Glucose would not be able to get into the cell treated with this chemical
Potassium mwill not be able to move or transport may be affected
Explanation:
As all membrane transport proteins are inactivated glucose is abig molecule it cannot pass without transporter protein.
Potassium transport is through sodium potassium pump and leak channels. As all transport protein are affected so it should not be able to move but if drug does not affect them then they will be unaffected.
Answer:
1= weather
2= climate
3=climate
4= climate, i think
5= weather
6= weather
7= climate
8=weather
9= weather
10= weather
hope this helps!
Explanation:
weather is typically over a short period, climate is over a long time.
Answer:
b
Explanation:
stable climate will mean there is one climate and we will have the same animals and plants everywere
3. Other names for S- waves are secondary waves, shear waves, and sometimes elastic S-waves. Other names for P-waves are primary waves and compressional waves.
4. You need 3 stations, because scientists find the difference between the arrival times of the primary and the secondary waves at each of the 3 stations, then the time difference is used to determine the distance of the epicentre from each station. The greater the difference in time, the further away the epicentre is. A circle is drawn around each station, with a radius corresponding to the epicentre’s distance from that station. The point where the three circles meet is the epicentre. If you only had two stations, you could only predict the epicentre, as the point where all three circles meet wouldn’t be complete, you’d have to try and estimate where the third one would intercept. This would greaten the chance of error and isn’t as accurate.
Hope this helps!
Answer:
- final temperature (T2) = 748.66 K
- ΔU = w = 5620.26 J
- ΔH = 9367.047 J
- q = 0
Explanation:
ideal gas:
reversible adiabatic compression:
∴ q = 0
∴ w = - PδV
⇒ δU = δw
⇒ CvδT = - PδV
ideal gas:
⇒ PδV + VδP = RδT
⇒ PδV = RδT - VδP = - CvδT
⇒ RδT - RTn/PδP = - CvδT
⇒ (R + Cv,m)∫δT/T = R∫δP/P
⇒ [(R + Cv,m)/R] Ln (T2/T1) = Ln (P2/P1) = Ln (1 E6/1 E5) = 2.303
∴ (R + Cv,m)/R = (R + (3/2)R)/R = 5/2R/R = 2.5
⇒ Ln(T2/T1) = 2.303 / 2.5 = 0.9212
⇒ T2/T1 = 2.512
∴ T1 = 298 K
⇒ T2 = (298 K)×(2.512)
⇒ T2 = 748.66 K
⇒ ΔU = Cv,mΔT
⇒ ΔU = (3/2)R(748.66 - 298)
∴ R = 8.314 J/K.mol
⇒ ΔU = 5620.26 J
⇒ w = 5620.26 J
⇒ ΔH = ΔU + nRΔT
⇒ ΔH = 5620.26 J + (1 mol)(8.314 J/K.mol)(450.66 K)
⇒ ΔH = 5620.26 J + 3746.787 J
⇒ ΔH = 9367.047 J
