Answer: The final concentration of aluminum cation is 0.335 M.
Explanation:
Given: 
= 47.8 mL (1 mL = 0.001 L) = 0.0478 L
= 0.321 M, 
= 21.8 mL = 0.0218 L, 
= 0.366 M
As concentration of a substance is the moles of solute divided by volume of solution.
Hence, concentration of aluminum cation is calculated as follows.
![[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}](https://tex.z-dn.net/?f=%5BAl%5E%7B3%2B%7D%5D%20%3D%20%5Cfrac%7BM_%7B1%7DV_%7B1%7D%20%2B%20M_%7B2%7DV_%7B2%7D%7D%7BV_%7B1%7D%20%2B%20V_%7B2%7D%7D)
Substitute the values into above formula as follows.
![[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}\\= \frac{0.321 M \times 0.0478 L + 0.366 M \times 0.0218 L}{0.0478 L + 0.0218 L}\\= \frac{0.0153438 + 0.0079788}{0.0696}\\= 0.335 M](https://tex.z-dn.net/?f=%5BAl%5E%7B3%2B%7D%5D%20%3D%20%5Cfrac%7BM_%7B1%7DV_%7B1%7D%20%2B%20M_%7B2%7DV_%7B2%7D%7D%7BV_%7B1%7D%20%2B%20V_%7B2%7D%7D%5C%5C%3D%20%5Cfrac%7B0.321%20M%20%5Ctimes%200.0478%20L%20%2B%200.366%20M%20%5Ctimes%200.0218%20L%7D%7B0.0478%20L%20%2B%200.0218%20L%7D%5C%5C%3D%20%5Cfrac%7B0.0153438%20%2B%200.0079788%7D%7B0.0696%7D%5C%5C%3D%200.335%20M)
Thus, we can conclude that the final concentration of aluminum cation is 0.335 M.
Density (p) is defined as the mass (m) per unit volume (v) or:
p = m/v
Using this relationship, the volume is:
v = m/p
Using the given values of mass of 80 grams and density of 8 g/cm3, the sample volume is:
v = 80 grams/8 grams/cm3
v = 10 cm3
The final answer is 10 cm3.
The difference between ergosterol and cholesterol is much more obvious in their interactions with POPC and DOPC. Whereas cholesterol induces a strong condensing effect that thickens both POPC and DOPC bilayers, ergosterol shows no condensing effect in POPC and DOPC at all.
Answer:
a. Concave down
Linear increasing
b. Increases the reaction rate
c. The reaction approaches the saturation point of the enzyme
Explanation:
a. For the reaction with enzyme, the shape is concave down. The action of the enzyme on the preferred substrate is initially very rapid and decreases as the enzyme becomes saturated and the ratio of products to substrate increases to approach an equilibrium rate of reaction
For the reaction without enzyme, the shape is linear and increasing. Increase in the concentration of the substrate will increase the number of effective collisions that lead into product formation leading to an increased rate of the chemical reaction
b. The enzyme increases the proportion of effective combination of substrates to form the products
c. The curve of the reaction with enzyme flattens out because as the concentration of the substrate increases while that of the enzyme remains the same, the enzyme becomes saturated and less able to increase the rate of the reaction of the excess substrate.
