Answer:
second law of thermodynamics.
Explanation:
The second law of thermodynamics deals with interconversion of energy from one form to another. Although energy can be converted from one form to another, this conversion is never 100% efficient because energy is lost in certain ways such as through heat. In a combustion engine, it is not possible to recover the energy from the gasoline 100% since energy must be lost along the way via such means as heat losses. Hence I will be skeptical about such an advert.
Answer:
The correct option is (d).
Explanation:
It is given that,
1$ = 1500 pesos
We need to convert 360 pesos into dimes
We can convert 360 pesos to dollars as follows:

360 pesos is equal to $0.24
Also, 1 dollar = 10 dimes
We can covert 0.24 dollar to dimes as follows :
0.24 dollar = 10 × 0.24 dimes
0.24 dollar = 2.4 dimes
or
360 pesos = 2.4 dimes
Answer:
AB
Explanation:
When combining two elements in different groups, you would just put them together unless you have two or more of each element
Answer:
Hydrogen (H) Sulfur(S) and Oxygen(O)
Explanation:
Answer:
4.77 is the pH of the given buffer .
Explanation:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
![pH=-\log[K_a]+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BK_a%5D%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
![pH=-\log[K_a]+\log(\frac{[CH_3CH_2COONa]}{[CH_3CH_2COOH]})](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BK_a%5D%2B%5Clog%28%5Cfrac%7B%5BCH_3CH_2COONa%5D%7D%7B%5BCH_3CH_2COOH%5D%7D%29)
We are given:
= Dissociation constant of propanoic acid = 
![[CH_3CH_2COONa]=0.254 M](https://tex.z-dn.net/?f=%5BCH_3CH_2COONa%5D%3D0.254%20M)
![[CH_3CH_2COOH]=0.329 M](https://tex.z-dn.net/?f=%5BCH_3CH_2COOH%5D%3D0.329%20M)
pH = ?
Putting values in above equation, we get:
![pH=-\log[1.3\times 10^{-5}]+\log(\frac{[0.254 M]}{[0.329]})](https://tex.z-dn.net/?f=pH%3D-%5Clog%5B1.3%5Ctimes%2010%5E%7B-5%7D%5D%2B%5Clog%28%5Cfrac%7B%5B0.254%20M%5D%7D%7B%5B0.329%5D%7D%29)
pH = 4.77
4.77 is the pH of the given buffer .