The answer to this question would be: 3.125%
Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
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.
Answer:
ΔU=-369.2 kJ/mol.
Explanation:
We start from the equation:
Δ(H)=ΔU+Δ(PV), which is an extension of the well known relation: H=U+PV.
If Δ(PV) were calculated by ideal gas law,
PV=nRT
Δ(PV)=RTΔn.
Where Δn is the change of moles due to the reaction; but, this reaction does not give a moles change (Four moles of HCl produced from 4 moles of reactants), so Δ(PV)=0.
So, for this case, ΔH=ΔU.
The enthalpy of reaction given is for one mole of reactant, so the enthalpy of reaction for the reaction of interest must be multiplied by two:

ΔU=-369.2 kJ/mol.
Answer: The molecular formula will be 
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C= 70.6 g
Mass of H = 5.9 g
Mass of O = 23.5 g
Step 1 : convert given masses into moles.
Moles of C =
Moles of H =
Moles of O =
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = 
For H = 
For O =
The ratio of C : H: O= 4: 4:1
Hence the empirical formula is 
The empirical weight of
= 4(12)+4(1)+1(16)= 68g.
The molecular weight = 136 g/mole
Now we have to calculate the molecular formula.

The molecular formula will be=
Answer: 
Explanation:
The Chemical equation for the formation of rust is:
Iron + Water + Oxygen ----> Rust
4 Fe(s) + 6 H2O(l) + 3 O2(g) → 4 Fe(OH)3(s)
The Iron Hydroxide However dehydrates to produce Fe2O3 * nH2O