The cytoplasm slides and forms a pseudopodium in front to move the cell forwards.
Answers and Explanation:
a)- The chemical equation for the corresponden equilibrium of Ka1 is:
2. HNO2(aq)⇌H+(aq)+NO−2
Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.
b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:
ΔG= ΔGº + RT ln Q
Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)
At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:
⇒ 0 = ΔGº + RT ln Ka
ΔGº= - RT ln Ka
ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)
ΔGº= 19092.8 J/mol
c)- According to the previous demonstation, at equilibrium ΔG= 0.
d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:
ΔG= ΔGº + RT ln Q
Q= ((H⁺) (NO₂⁻))/(HNO₂)
Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)
Q= 1.88 10⁻⁴
We know that ΔGº= 19092.8 J/mol, so:
ΔG= ΔGº + RT ln Q
ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)
ΔG= -2162.4 J/mol
Notice that ΔG<0, so the process is spontaneous in that direction.
Answer:
Below!
Explanation:
A planet with oceans, continents, and plate tectonics maximizes opportunities for speciation and natural selection, whereas a similar planet without plate tectonics provides fewer such opportunities. Plate tectonics exerts environmental pressures that drive evolution without being capable of extinguishing all life.
Answer: The density of 0.50 grams of gaseous carbon stored under 1.50 atm of pressure at a temperature of -20.0 °C is 0.867 g/L.
Explanation:
- d = m/V, where d is the density, m is the mass and V is the volume.
- We have the mass m = 0.50 g, so we must get the volume V.
- To get the volume of a gas, we apply the general gas law PV = nRT
P is the pressure in atm (P = 1.5 atm)
V is the volume in L (V = ??? L)
n is the number of moles in mole, n = m/Atomic mass, n = 0.50/12.0 = 0.416 mole.
R is the general gas constant (R = 0.082 L.atm/mol.K).
T is the temperature in K (T(K) = T(°C) + 273 = -20.0 + 273 = 253 K).
- Then, V = nRT/P = (0.416 mol)(0.082 L.atm/mol.K)(253 K) / (1.5 atm) = 0.576 L.
- Now, we can obtain the density; d = m/V = (0.50 g) / (0.576 L) = 0.867 g/L.
