Answer:
= 1.5 eq
Explanation:
One definition of an equivalent weight is that it is mass of a substance that gains or loses 1 mole of electrons.
Al3+ has lost 3 e-, so there are 3 equivalent weights in 1 mol Al3+.
1 mol Al3+ =3 eq. wts.
1 mol Al x(27 g / 1 mol)x(1 mol / 3 eq. wts.) = 9.0 g = 1 eq. wts.
13.5 g Al3 + x (1 eq.wt. / 9.0 g) = 1.5 eq
Answer is: 0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K.
Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - <span>Henry's law constant.
</span>c - solubility of a gas at a fixed temperature in a particular solvent.
c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.
Answer:
0.7μM = 0.6 μM = 0.5 μM > 0.4 μM > 0.3 μM > 0.2 μM
Explanation:
An enzyme solution is saturated when all the active sites of the enzyme molecule are full. When an enzyme solution is saturated, the reaction is occurring at the maximum rate.
From the given information, an enzyme concentration of 1.0 μM Y can convert a maximum of 0.5 μM AB to the products A and B per second means that a 1.0 M Y solution is saturated when an AB concentration of 0.5 M or greater is present.
The addition of more substrate to a solution that contains the enzyme required for its catalysis will generally increase the rate of the reaction. However, if the enzyme is saturated with substrate, the addition of more substrate will have no effect on the rate of reaction.
<em>Therefore the reaction rates at substrate concentrations of 0.7μM, 0.6 μM, and 0.5 μM are equal. But the reaction rate at substrate concentrations of 0.2 μM is lower than at 0.3 μM, 0.3 μM is lower than 0.4 μM and 0.4 μM is lower than 0.5 μM, 0.6 μM and 0.7 μM.</em>
B is the answer to ur question
