Answer:
d, 40 dm3.
Explanation:
According to Avogadro's law, the mole ratio of chemicals in a reaction is equal to the ratio of volumes of chemicals reacted (for gas).
From the equation, the mole ratio of N2 : H2 : NH3 = 1 : 3 : 2, meaning 1 mole of N2 reacts completely with 3 moles of H2 to give 2 moles of NH3, the ratio of volume required is also equal to 1 : 3 : 2.
Considering both N2 and H2 have 30dm3 of volume, but 1 mole of N2 reacts completely with 3 moles of H2, so we can see H2 is limiting while N2 is in excess. Using the ratio, we can deduce that 10dm3 equals to 1 in ratio (because 3 moles ratio = 30dm3).
With that being said, all H2 has reacted, meaning there's no volume of H2 left. 2 moles of NH3 is produced, meaning the volume of NH3 produced = 10 x 2 = 20 dm3. (using the ratio again)
1 mole of N2 has reacted, meaning from the 30dm3, only 10 dm3 has reacted. This also indicate that 20 dm3 of N2 has not been reacted.
So at the end, the mixture contains 20dm3 of NH3, and 20 dm3 of unreacted N2. Hence, the answer is d, 40 dm3.
Answer:
Explanation:
Hello,
Based on the given undergoing chemical reaction is is rewritten below:
By stoichiometry we find the minimum mass of H2SO4 (in g) as shown below:
Moreover, mass of H2 gas (in g) would be produced by the complete reaction of the aluminum block turns out:
Best regards.
Answer:
11.3 g.
Explanation:
Hello there!
In this case, since the combustion of butane is:
Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:
Therefore, the resulting mass of water is:
Best regards!
The volume of the dry gas at stp is calculated as follows
calculate the number on moles by use of PV =nRT where n is the number of moles
n is therefore = Pv/RT
P = 0.930 atm
R(gas contant= 0.0821 L.atm/k.mol
V= 93ml to liters = 93/1000= 0.093L
T= 10 + 273.15 = 283.15k
n= (0.930 x0.093) /(0.0821 x283.15) = 3. 72 x10^-3 moles
At STp 1 mole = 22.4L
what about 3.72 x10^-3 moles
by cross multiplication
volume = (3.72 x10^-3)mole x 22.4L/ 1 moles = 0.083 L or 83.3 Ml
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.
