Answer:
Was the metal oxidized?
Cu = No | No | No
Fe = Yes | No | No
Zn = Yes | Yes | No
Oxidizing Agent:
Cu = None | None | None
Fe = Cu, NO3 | None | None
Zn = Cu, NO3 | Fe, NO3 | None
Copper has a FCC i.e. face centered cubic crystal structure. The 100 plane is essentially a planar section of the cubic cell where 4 Cu atoms occupy the 4 corners of the plane along with 1 Cu atom at the center of that plane. Each of the Cu atoms in the corners is shared by 4 adjacent unit cells. Thus, there are 2 Cu atoms present in the 100 plane (4*1/4 + 1 = 2).
Now, the planar density PD along the 100 plane is given as:
PD(100) = # atoms in the 100 plane/Area of 100 plane
= 
Here R = radius = 0.128 nm = 
PD = 
Answer:
3.02e23
Explanation:
carbon molecular weight = 12 g /mol
1 mol = 6.02e23 atoms
your sample, 6 g / 12 g/mol = 0.5 mol
0.5 x 6.02e23 =
Primary alkyl halides tend to undergo the SN2 reaction mechanism in nucleophilic substitution since there is less steric hindrance for nucleophilic attack and the carbocations that they form are not as stable as those formed from tertiary alkyl halides.
1-bromopentane > 1-bromo 2-methylbutane > <span>1-bromo-3-methylbutane</span>> 2-bromo 2-methylbutane
Answer:
(a) 13.7 g.
(b) 28.91 g.
Explanation:
- molality (m) is the no. of moles of solute dissolved in 1.0 kg of solvent.
∴ m = (no. of moles of solute)/(mass of water (kg))
<em>∴ m = (mass/molar mass of solute)/(mass of water (kg)).</em>
<em />
<u><em>(a) Calculate the mass of CaCl₂·6H₂O needed to prepare 0.125 m CaCl₂(aq) by using 500. g of water.</em></u>
∵ m = (mass/molar mass of CaCl₂·6H₂O)/(mass of water (kg)).
m = 0.125 m, molar mass of CaCl₂·6H₂O = 219.0757 g/mol, mass of water = 500.0 g = 0.5 kg.
∴ 0.125 m = (mass of CaCl₂·6H₂O / 219.0757 g/mol)/(0.5 kg).
∴ mass of CaCl₂·6H₂O = (0.125 m)(219.0757 g/mol)(0.5 kg) = 13.7 g.
<u><em>(b) What mass of NiSO₄·6H₂O must be dissolved in 500. g of water to produce 0.22 m NiSO₄(aq)?</em></u>
∵ m = (mass/molar mass of NiSO₄·6H₂O)/(mass of water (kg)).
m = 0.22 m, molar mass of NiSO₄·6H₂O = 262.84 g/mol, mass of water = 500.0 g = 0.5 kg.
∴ 0.125 m = (mass of NiSO₄·6H₂O / 262.84 g/mol)/(0.5 kg).
∴ mass of NiSO₄·6H₂O = (0.22 m)(262.84 g/mol)(0.5 kg) = 28.91 g.
