<span>0.0165 m
The balanced equation for the reaction is
AgNO3 + MgCl2 ==> AgCl + Mg(NO3)2
So it's obvious that for each Mg ion, you'll get 1 AgCl molecule as a product. Now calculate the molar mass of AgCl, starting with looking up the atomic weights.
Atomic weight silver = 107.8682
Atomic weight chlorine = 35.453
Molar mass AgCl = 107.8682 + 35.453 = 143.3212 g/mol
Now how many moles were produced?
0.1183 g / 143.3212 g/mol = 0.000825419 mol
So we had 0.000825419 moles of MgCl2 in the sample of 50.0 ml. Since concentration is defined as moles per liter, do the division.
0.000825419 / 0.0500 = 0.016508374 mol/L = 0.016508374 m
Rounding to 3 significant figures gives 0.0165 m</span>
Answer:
G]ns^2np^5 group 17 (p-block)
G]ns^2np^2 group 14 (p-block)
G]ns^2mf^14 group 16 (f-block)
Explanation:
The outermost electronic configuration of an element shows the group to which it belongs in the periodic table as shown above in the answer. In addition, to that, we can be able to know from its electronic configuration, whether the element is a metal or not.
For instance;
G]ns^2mf^14 is a rare earth metal, G]ns^2np^2 group 14 is a metalloid while G]ns^2np^5 group 17 is a nonmetal.
Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen 
if it occupies 31.6 mL at 
.
Explanation:
Given : Mass of oxygen = 0.023 g
Volume = 31.6 mL
Convert mL into L as follows.
Temperature = 
As molar mass of 
is 32 g/mol. Hence, the number of moles of are calculated as follows.
Using the ideal gas equation calculate the pressure exerted by given gas as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the value into above formula as follows.
Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen if it occupies 31.6 mL at .
Answer:
Iron (II) nitrate is ionic compound
Explanation:
Formula:
Fe(NO₃)₂
Iron (II) nitrate is ionic compound.
Its molecular mass is 179.85 g/mol.
NO⁻₃ is anion while Fe⁺² is cation.
Iron loses its two electron which is accepted by nitrate.
Its molecular formula can be written as FeN₂O₆.
Its color is pale green.
Its melting point is 333.65 K.
It is para magnetic compound.
it is mostly present in non hydrated form.
It is also known as ferric nitrate.
It is used to form sodium amide.
Its is also used catalyst.
