They are the outer layer of the electron layers.
<span>The student is incorrect because helium has 2 valence electrons and it's in group 18 because the first energy level is full. Although helium is placed in Group 18 which generally has 8 valence electrons, it does not have 8 valence electrons as the student suggested. It was grouped together with the noble gases because it exhibits similar properties with them. </span>
Answer:
Enzyme
Explanation:
An enzyme is an organic catalyst that speeds up a reaction but can be recovered unchanged.
Human lactase consists of 4092 amino acid groups and has a molar mass of about 280 000 u, so it has a complex structure.
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
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
