Answer:
The answer is 465.6 mg of MgI₂ to be added.
Explanation:
We find the mole of ion I⁻ in the final solution
C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol
But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.
So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.
Hence, the weight of MgI₂ must be added is
Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.
Answer:
Molarity is 0.04M
Explanation:
First of all, let's determinate the moles of aspirin in that sample
Mass / Molar mass = Moles
360 mg = 0.360 g
0.360 g / 180 g/m = 0.002 moles
This moles that are included in 200 mL of solution, are also in 50 mL.
So molarity is mol/L
50 mL = 0.05 L
0.002 m / 0.05 L = 0.04M
Cadmium chloride is a highly soluble compound. The equation for its dissolution is:
CdCl₂(s) → Cd⁺²(aq) + 2Cl⁻(aq)
This dissociation in water allows for the cadmium and chlorine ions to take part in reactions. This is the reason that solutions of chemicals are prepared when a reaction needs to take place.
Answer:
86 mL
Explanation:
First find the moles of Pb (NO3)2
n=cv
where
c ( concentration)= 0.210 M
v ( volume in L) =0.05
n= 0.210 × 0.05
n= 0.0105
Using the mole ratio, we can find the moles of KCl by multiplying by 2
n (KCl) =0.0105 ×2
=0.021
v (KCl)= n/ c
= 0.021/ 0.244
=0.08606557377
=0.086 L
= 86 mL
