Answer:
The significance of "Er" in the diagram is :
B.) Threshold energy for reaction
Explanation:
Threshold energy : It is total amount of energy required by the reactant molecule to reach the transition state .
Activation energy : It is the excess energy absorbed by the molecules to reach the transition state.
<u>Activation Energy = Threshold Energy - Average Kinetic Energy</u>
<u>This means Activation energy decreases on increasing kinetic energy</u>
On increasing Temperature average kinetic energy of the molecule increases which reduces the activation energy and the reaction occur faster in that case.
Catalyst also reduces the Activation energy.
<u>Er = Threshshold energy for reaction at 30 degree</u>
<u>Ea = Activation Energy</u>
<u>The given figure shows that the threshold energy decreases on increasing the temperature</u>
<u>Only the molecule having energy greater than Er can react to form product</u>
Answer: 72 grams of 
are needed to completely burn 19.7 g 
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 
of particles.
To calculate the number of moles, we use the equation:
Putting in the values we get:
According to stoichiometry:
1 mole of 
requires 5 moles of oxygen
0.45 moles of require= 
moles of oxygen
Mass of 
72 grams of are needed to completely burn 19.7 g
Answer:
V = 12.93 L
Explanation:
Given data:
Number of moles = 0.785 mol
Pressure of balloon = 1.5 atm
Temperature = 301 K
Volume of balloon = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will put the values.
V = nRT/P
V = 0.785 mol × 0.0821 atm.L/ mol.K × 301 K / 1.5 atm
V = 19.4 L /1.5
V = 12.93 L
Answer:
Explanation:
use the equation
moles = mass/mr
=19.9/79.5
=0.250moles of CuO
then do the same for
H = 2.02/1
=2.02
so CuO is the limiting reagent because there is less amount of it.
Hope this helps :)
Answer:
Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
![pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
Regards.
