Explanation:
There are two aspects of a reaction:
Thermodynamics:
Combustion of the compound is reaction with oxygen.
The carbohydrates contain more content of oxygen as compared to fats.
Hence, carbohydrates have lot of oxygen content which are already partially oxidized as compared to fats. Hence, combustion of the carbohydrates is a faster process.
Kinetics:
The molecular reacts also shows that the combustion of the carbohydrates is a faster process.
Answer:
More energy is released from the old substance than the new substance needs to form its chemical bonds
Explanation:
Edge 2020
Answer: The mass of the sample will be 1417.7 grams.
Explanation:
We are given:
This means that 1 mole of NaCl has an enthalpy of fusion of 30.2 kJ
1 mole of NaCl has a mass of 58.44 grams.
So, 30.2 kJ of heat is require for a mass 58.44 grams of NaCl
So, 732.6 kJ of heat will be required for = 
= 1417.65 grams of NaCl.
Hence, the mass of NaCl sample will be 1417.7 grams.
Answer:
a. pH = 2 b. pH = 3 c. pH = 1 d. Unanswerable
Explanation:
pH = -log[H+] OR pH = -log{H3O+]
and inversely
pOH = -log[OH-]
1. Determine what substance you are working with, (acid/base)
2. Determine whether or not that acid or base is strong or weak.
a. 1.0 x 10^-2M HCl
HCl is a strong acid, therefore it will dissociate completely into H+ and Cl- with all ions going to the H+, therefore, the concentration of HCl and concentration of H+ are going to be equal, meaning we simply take the negative logarithm of the concentration of HCl and that would equal pH
pH = -log[H+]
pH = -log(1.0x10^-2)
pH = 2
b. 1.0 x 10^-3M HNO3
HNO3 like part a, is a strong acid, therefore it would simply require you to take the negative logarithm of the concentration of the compound itself, to find its pH.
pH = -log[H+]
pH = -log(1.0 x 10^-3)
pH = 3
c. 1.0 x 10^-1M HI
Like the previous parts, HI is a strong acid
pH = -log[H+]
pH = -log(0.10)
pH = 1
d. HB isn't an element, nor is it a compound so that would be unanswerable.
Answer:
a) 90 kg
b) 68.4 kg
c) 0 kg/L
Explanation:
Mass balance:
w is the mass flow
m is the mass of salt
v is the volume flow
C is the concentration
![-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)](https://tex.z-dn.net/?f=-%5Bln%282000L%2B3%2AL%2Fmin%2At%29-ln%282000L%29%5D%3Dln%28m%29-ln%2890kg%29)
![-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)](https://tex.z-dn.net/?f=-ln%5B%282000L%2B3%2AL%2Fmin%2At%29%2F2000L%5D%3Dln%28m%2F90kg%29)
![m=90kg*[2000L/(2000L+3*L/min*t)]](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2At%29%5D)
a) Initially: t=0
![m=90kg*[2000L/(2000L+3*L/min*0)]=90kg](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2A0%29%5D%3D90kg)
b) t=210 min (3.5 hr)
![m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2A210min%29%5D%3D68.4kg)
c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.
