The volume of 0.160 m Li2S solution required to completely react with 130 ml of 0.160 CO(NO3)2 is calculated as below
write the reacting equation
Co(NO3)2 + Li2S = 2LiNO3 + COS
find the moles of CO(NO3)2 = molarity x volume
= 130 ml x 0.160=20.8 moles
since the reacting moles between CO(NO3)2 to LiS is 1:1 the moles of LiS is also 20.8 moles
volume of Lis is therefore = moles of Lis/ molarity of LiS
= 20.8/0.160 = 130 Ml
Your answer is D. 8
16 = 2^4
72 = 2^3*3^2
So you'll choose 2^3 = 8
1 mole has 6.02*10^23 molecules in it.
1 nickel (II) chloride molecule, NiCl2, has 1 Ni atom in it.
so 1 mole of nickel (II) chloride molecule has 1 mole of Ni atom in it.
so 100 moles of nickel (II) chloride molecule has 100*6.02*10^23
= 6.02*10^25 Ni atom in it.
Vs = 1.0 mL = 0.001 L
c((NH4)2CO3) = <span>0.02 M
n(</span>(NH4)2CO3) = ?
For the purpose, here we will use the next equation:
c=n/V ⇒ n=cxV
n((NH4)2CO3) = 0.02M x 0.001L
n((NH4)2CO3) = 2x10⁻⁵ mole of (NH4)2CO3 is presented in the solution
