Answer: d. More than 6.5 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.
Explanation: 
As can be seen from the chemical equation, 2 moles of aluminium react with 3 moles of copper chloride.
According to mole concept, 1 mole of every substance weighs equal to its molar mass.
Aluminium is the limiting reagent as it limits the formation of product and copper chloride is the excess reagent as (14-7.5)=6.5 g is left as such.
Thus 54 g of of aluminium react with 270 g of copper chloride.
1.50 g of aluminium react with=
of copper chloride.
3 moles of copper chloride gives 3 moles of copper.
7.5 g of copper chloride gives 7.5 g of copper.
Answer:
III, IV, and V
Explanation:
The complex [CO(NH3)6]3+ is a diamagnetic complex. It a low spin d^6 complex. Most d^6 complexes are low spin due to the higher crystal field stabilization energy of the low spin over the high spin arrangement.
d^6 metal complexes are known to be octahedral (a coordination number of 6 leads to octahedral geometry). Octahedral complexes does not have geometric isomers rather, may exist as the fac or her stereo isomers.
Given: C= 81.70% = 81.70g
H = 18.29% = 18.29g
<span>The number of moles is given by: n= Given mass (m)/Molar Mass (M)
</span>M of C = 12 g/mol
M of H = 1 g/mol
Thus, the number of moles of carbon = 81.70g / 12gmol= 6.83moles
and, the number of moles of hydrogen = 18.29/1g/mol = 18.29 moles
The ration of C moles with hydrogen :
H:C = 18.29moles/6.83moles= 2.67 ≈3
Thus, the empirical formula is C3H8
Answer:
The mass of nickel is 48μg
Explanation:
Parts per billion is a way to describe small concentrations and is defined as the ratio between μg of solute and kg of solvent.
If a solution of nickel in propanol is 20ppb, contains 20μg of nickel in 1 kg of propanol.
Thus, a sample of 2.4kg of propanol will contain:
2.4kg × (20μg nickel / 1kg) = 48μg nickel
<h3>The mass of nickel is 48μg</h3>
Answer:
NaI > Na2SO4 > Co Br3
meaning that NaI has the highest freezing point, and Co Br3 has the lowest freezing point.
Explanation:
The freezing point depression is a colligative property.
That means that it depends on the number of solute particles dissolved.
The formula to calculate the freezing point depression of a solution of a non volatile solute is:
ΔTf = i * Kf * m
Where kf is a constant, m is the molality and i is the van't Hoff factor.
Molality, which is number of moles per kg of solvent, counts for the number of moles dissolved and the van't Hoff factor multipllies according for molecules that dissociate.
The higher the number of molecules that dissociate, the higher the van't Hoff, the greater the freezing point depression and the lower the freezing point.
As the question states that you assume equal concentrations (molality) and complete dissociation you just must find the number of ions generated by each solute, in this way:
NH4 I → NH4(+) + I(-) => 2 ions
Co Br3 → Co(+) + 3 Br(-) => 4 ions
Na2SO4 → 2Na(+) + SO4(2-) => 3 ions.
So, Co Br3 is the solute that generate more particles and that solution will exhibit the lowest freezing point among the options given, Na2SO4 is next and the NaI is the third. Ordering the freezing point from higher to lower the rank is NaI > Na2SO4 > CoBr3, which is the answer given.
