Metals naturally form cations.
because it can influence how frequently and sufficiently the particles collide depending on the space it has to do so, for example a large surface area would be have a slower rate of reaction and a lower temperature. (the rate of reaction in terms of concentration, it is diffused from high to low)
Answer:
The name of this compound is :
Bi2(CO3)3 = Bismuth Carbonate
Explanation:
The name of the compound is derived from the name of the elements present in it.
The rule followed while naming the compound are:
1. The first element (always the cation) is named as such .
2. The second element (The anion) end with "-ate , -ide ," etc
3. NO prefix is added while naming the first element.
For example : Bi2 can't be named as Dibismuth
Na2 = Can't be named as disodium
Hence the compound :
Bi2(CO3)3 contain two element : Bi and CO3. Here , Bi = cation (named as such) and CO3 = anion (named according to rules)
Bi = Bismuth
CO3 = carbonate
Bi2(CO3)3 = Bismuth Carbonate
The molecular mass of this compound is :
Molecular mass = 2 (mass of Bi) + 3(mass of C) + 6(mass of O)
= 2 (208.98)+3(12.01)+6(15.99)
= 597.987 u
We will assume that the solvent is water. So, if we have 100 grams of the solution, 19 grams will be sodium hydroxide, while the remaining 81 grams will be water.
The molar weight of sodium hydroxide, NaOH, is 40. The molar weight of water is 18. Finding the moles of each:
NaOH:
19 / 40 = 0.475
Water:
81 / 18 = 4.5
Total moles present:
4.5 + 0.475 = 4.975 moles
The mole fraction of NaOH is:
0.475 / 4.975 = 0.0955
The mole fraction of NaOH is 0.0955
Answer: 36.53g
Explanation:
First we need to find the amount of NaCl that dissolves in 1L of the solution that produced 5M of NaCl
Molarity = 5M
MM of NaCl = 58.45
Molarity = Mass conc (g/L) / MM
Mass conc. (g/L) of NaCl = Molarity x MM
= 5 x 58.45 = 292.25g
Next, we need to find the amount that will dissolve in 125mL(i.e 0.125L)
From the calculations above,
292.25g of NaCl dissolved in 1L
Therefore Xg of NaCl will dissolve in 0.125L of the solution i.e
Xg of NaCl = 292.25 x 0.125 = 36.53g.
Therefore 36.53g of NaCl will dissolve in 125mL of the solution
