The area where the periodic table has the highest electronegativity are Groups 1 and Groups 7 because they are close on giving away an electron and the other is close to becoming stable.
Answer:
KBr is limiting reactant.
Explanation:
Given data:
Mass of KBr =4g
Mass of Cl₂ = 6 g
Limiting reactant = ?
Solution:
Chemical equation:
2KBr + Cl₂ → 2KCl + Br₂
Number of moles of KBr:
Number of moles = mass/molar mass
Number of moles = 4 g/ 119 gmol
Number of moles = 0.03 mol
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 6 g/ 70 gmol
Number of moles = 0.09 mol
Now we will compare the moles of reactant with product.
KBr : KCl
2 : 2
0.03 : 0.03
KBr : Br₂
2 : 1
0.03 : 1/2×0.03= 0.015
Cl₂ : KCl
1 : 2
0.09 : 2/1×0.09 = 0.18
Cl₂ : Br₂
1 : 1
0.09 : 0.09
Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂ is present in excess.
Answer:
Instrumental methods
surface analysis
high performance liquid chromatography
atomic spectroscopy
potentiometry
Classical methods
precipitation titration
gravimetric analysis
Explanation:
Instrumental methods of analysis are those analytical methods in which the responsibility of detection has been removed from human beings and placed on automated instruments while classical methods are those analytical methods in which the responsibility of detection remains the responsibility of human beings.
Many instrumental methods such as HPLC rely on computer screens as readout devices.
Answer:
114.7g
Explanation:
Given parameter:
Mass of Na = 45g
Unknown:
Mass of produced NaCl = ?
Solution:
The reaction equation is shown below;
2Na + Cl₂ → 2NaCl
Now, we solve from the known to the unknown. The known here is the mass of Na and it is sufficient to solve the problem;
Number of moles = 
Molar mass of Na = 23g/mol
Number of moles = 
= 1.96moles
2 mole of Na produced 2 moles of NaCl
1.96 moles of Na will produce 1.96 moles of NaCl
Mass of NaCl = number of moles x molar mass
= 1.96 x [23 + 35.5]
= 114.7g
