Answer:
a. neutral
b. salts
c. salt
Explanation:
Organic salts are a dense number of ionic compounds with innumerable characteristics. They are previously derived from an organic compound, which has undergone a transformation that allows it to be a carrier of a charge, and that in addition, its chemical identity depends on the associated ion.
Organic salts are usually stronger acids or bases than inorganic salts. This is because, for example, in the amine salts, it has a positive charge due to its bond with an additional hydrogen: A + -H. Then, in contact with a base, donate the proton to be a neutral compound again
RA + H + B => RA + HB
H belongs to A, but it is written as it is involved in the neutralization reaction.
On the other hand, RA + can be a large molecule, unable to form solids with a crystalline network stable enough with the hydroxyl anion or oxyhydrile OH–.
When this is so, salt RA + OH– behaves as a strong base; even as basic as NaOH or KOH
Salt makes the freezing point of water decrease, so it would freeze at a warmer temperature than regular water
1:2
The ratio of carbon, hydrogen, and oxygen in most carbohydrates is 1:2:1. This means for every one carbon atom there are two hydrogen atoms and one...
Answer:
The answer to your question is: 2.20 x 10 ²³ molecules
Explanation:
Data
mass = 45.7 g
molecules of CF₂Cl₂ = ?
Process
1.- Calculate the mass number of CF₂Cl₂
C = 12 F =2 x 19 Cl = 2 x 35.5
total = 12 + 38 + 71
total = 121 g
2.- Use the Avogradro's number to solve the problem
121 g ------------------- 6.023 x 10²³ molecules of CF₂Cl₂
45.7 g --------------- x
x = (45.7 x 6.023 x 10²³) / 121
x = 2.75 x 10²⁵ / 121
x = 2.20 x 10 ²³ molecules
We can use combined gas laws to solve for the volume of the gas
where P - pressure, V - volume , T - temperature and k - constant
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
T1 - temperature in Kelvin - 20 °C + 273 = 293 K
T2 - 40 °C + 273 = 313 K
substituting the values
V = 17.8 L
volume of the gas is 17.8 L
