Explanation:
To form bonds with noble gases, a lot of energy is required to form those bonds. Halogens, on the other hand, are extremely reactive. ... The halogens tend to be very reactive, while the noble gases are in no way reactive and don't bond easily, if at all.
Answer:
The answer to your question is E = 83.2 J
Explanation:
Data
Element: Gold
Initial temperature = T1 = 5°C
Final temperature = T2 = 37°C
mass = 20 g
Specific heat = 130 J/kg°K
Process
1.- Convert temperature to kelvin
T1 = 273 + 5 = 278°K
T2 = 273 + 37 = 310°K
2.- Convert mass to kg
1000 g --------------- 1 kg
20 g --------------- x
x = (20 x 1)/1000
x = 0.02 kg
3.- Formula
E = mC(T2 -T1)
4.- Substitution
E = (0.02)(130)(310 - 278)
E = (0.02)(130)(32)
E = 83.2 J
2H2 + O2 →2H2O
This is a limiting reactant problem in which you need to figure out which reactant is the limiting reactant. You can usually look at the information given and see that hydrogen has the smaller amount of mass used, but it is always good to check that assumption through stoichiometry:
(9.43g H2/2.02g/mol)(2mol H2O/2mol H2)(18.02gH2O/1mol H2O)= 84.1g H2O produced from 9.43g of H2
(12.98g O2/32.00g/mol O2)(2mol H2O/1mol O2)(18.02gH2O/1mol H2O)= 14.6g H2O
So my assumption that hydrogen makes the smaller amount of water is wrong since oxygen (it is the limiting reactant) created the smaller amount of water with 14.6g, so 14.6g is the maximum amount of water that can be formed
Answer:
The mass of NaBr needed is 0.22969 g.
Explanation:
1 mole of NaBr contains 22.4 dm^3 of NaBr
Therefore, 0.05 dm^3 (50/1000 = 0.05 dm^3) of NaBr = 0.05/22.4 = 0.00223 mol of NaBr
MW of NaBr = 23 + 80 = 103 g/mol
Mass of NaBr = number of moles × MW = 0.00223 × 103 = 0.22969 g
Solution:- Halogens are fluorine, chlorine, bromine, iodine and Astatine(F, Cl, Br, I and At) and these are present in the 7A group of the periodic table. In modern periodic table this group is written as 17th group. This group elements have 7 valence electrons and so they are in group 7A.
So, the right choice is D. 7A.