Answer : The expected coordination number of NaBr is, 6.
Explanation :
Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.
This is represented by,
When the radius ratio is greater than 0.155, then the compound will be stable.
Now we have to determine the radius ration for NaBr.
Given:
Radius of cation, 
= 102 pm
Radius of cation, 
= 196 pm
As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.
The relation between radius ratio and coordination number are shown below.
Therefore, the expected coordination number of NaBr is, 6.
Answer:
The enthalpy of the reaction is –184.6 kJ, and the reaction is exothermic.
Explanation:
1) Balanced chemical reaction:
2H2 + O2 -> 2H20
Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O
2) Reactant quantities converted to moles
H2: 5.00 g / 2 g/mol = 2.5 mol
O2: 50.0 g / 32 g/mol = 1.5625 mol
Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).
3) Products
H2 totally consumed -> 0 mol at the end
O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end
H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.
Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol
4) Pressure
Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K
p = nRT/V = 7.9 atm
Answer:
B) CaO(s) + H2O(l) --> Ca(OH)2(aq)
Explanation:
This is the only reaction with a negative enthalpy value. Exothermic reactions have a negative enthalpy.
Given: C= 81.70% = 81.70g
H = 18.29% = 18.29g
<span>The number of moles is given by: n= Given mass (m)/Molar Mass (M)
</span>M of C = 12 g/mol
M of H = 1 g/mol
Thus, the number of moles of carbon = 81.70g / 12gmol= 6.83moles
and, the number of moles of hydrogen = 18.29/1g/mol = 18.29 moles
The ration of C moles with hydrogen :
H:C = 18.29moles/6.83moles= 2.67 ≈3
Thus, the empirical formula is C3H8
