Answer:
1. 266.22 g/mol
2. 168.81 g/mol
3. 223.35 g/mol
4. 199.88 g/mol
Explanation:
For you to calculate the molar mass of the salt you need to sum the molar masses of every element in the salt.
In the first salt, PdBr
, the subscript 2 means that there are 2 atoms of Br. So for you to calculate the molar mass of the salt you need to sum the molar mass of Pd and 2 times the molar mass of Br, as follows:
106 g/mol + 2(79.90 g/mol) = 266.22 g/mol
In the second salt BeBr there are 2 atoms of Br and 1 of Be, so the molar mass is:
9.012 g/mol +2(79.90 g/mol) = 186.22 g/mol
In the third salt CuBr there are 2 atoms of Br and 1 of Cu, so the molar mass is:
63.55 g/mol + 2(79.90 g/mol) = 223.35 g/mol
And in the fourth salt CaBr there are 2 atoms of Br and 1 of Ca, so the molar mass is:
40.08 g/mol + 2(79.90 g/mol) = 199.88 g/mol
Answer:
Quantity of Carbon is 4.09 gm
Explanation:
Equation of carbon reacting with oxygen to give carbon dioxide is given by
C + 
⇒ C
One mole of carbon reacts with one mole of Oxygen in this reaction to give One mole of Carbon dioxide.
So, 12 gm of carbon reacts with 32 gm of Oxygen in this reaction to give 44 gm of carbon dioxide.
15 gm of C was formed in this reaction
Oxygen used in this reaction = 
×32 = 10.91 gm ,
Thus Oxygen is in sufficient quantity in the reaction.
Now,
Carbon that must be used = ×12 = 4.09 gm.
The answer is D because the air is made of nitrogen and oxygen. The reaction is endothermic.
Hope this helps you! :)
Answer:
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Explanation:
From the given information:
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.
In this same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;
Rate factor in the absence of catalyst:
Rate factor in the presence of catalyst:
Assuming the catalyzed reaction and the uncatalyzed reaction are taking place at the same temperature :
Then;
the ratio of the rate factors can be expressed as:
![\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}](https://tex.z-dn.net/?f=%5Cdfrac%7Bk_2%7D%7Bk_1%7D%3D%7B%20%20%5Cdfrac%20%7Be%5E%7B%5B%20%20Ea_1%20-%20Ea_2%20%5D%20%7D%7D%7BRT%7D%20%7D%7D)
Thus;
Let say the assumed temperature = 25° C
= (25+ 273)K
= 298 K
Then ;
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
