Ok to answer this question we firsst need to fin the number of mol of Urea (CH4N2O). to do this we simply :
1 mol of urea =15/60.055 = 0.25mol
therefore 200g of water contain 0.25mol
the next step is to determine the malality of our solution in 200g of water, to do this we say:
200 g = 1Kg/1000g = 0.2kg
therefor 0.25mol/0.2Kg = 1.25mol/kg
and from the equation:
we know that i = 1
we are given Kf
b is the molality that we just calculated
therefore;
the solutions freezing point is -2.325°C
The molecular weight of water is <span>18.01528 g/mol.
So in 2.92 grams there are 2.92/</span>18.01528 = 0.1621 mol of particles.
1 mol contains 6,02214 × 10^<span>23 particles by definition.
So the nr of H2O molecules is </span>0.1621 * 6,02214 × 10^23 = 0,9761 × 10^23.
Every molecule has 2 H atoms, so you have to double that.
2* 0,9761 × 10^23 = 1.952 × 10^23.
Answer:
The lock-and-key model:
c. Enzyme active site has a rigid structure complementary
The induced-fit model:
a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.
Common to both The lock-and-key model and The induced-fit model:
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.
d. Substrate binds to the enzyme through non-covalent interactions
Explanation:
Generally, the catalytic power of enzymes are due to transient covalent bonds formed between an enzyme's catalytic functional group and a substrate as well as non-covalent interactions between substrate and enzyme which lowers the activation energy of the reaction. This applies to both the lock-and-key model as well as induced-fit mode of enzyme catalysis.
The lock and key model of enzyme catalysis and specificity proposes that enzymes are structurally complementary to their substrates such that they fit like a lock and key. This complementary nature of the enzyme and its substrates ensures that only a substrate that is complementary to the enzyme's active site can bind to it for catalysis to proceed. this is known as the specificity of an enzyme to a particular substrate.
The induced-fit mode proposes that binding of substrate to the active site of an enzyme induces conformational changes in the enzyme which better positions various functional groups on the enzyme into the proper position to catalyse the reaction.
Hey there : !
density = 10.5 g/cm³
volume = 23.6 cm³
therefore:
D = m / V
10.5 = m / 23.6
m = 10.5 * 23.6
m = 247.8 g
hope this helps!
