Answer:
ρ = 4407.03 kg/m³
Explanation:
The Density of a metal alloy is given by the following equation:
1/ρ = m₁/ρ₁ + m₂/ρ₂ + m₃/ρ₃
where,
ρ = density of allow = ?
ρ₁ = density of Titanium (Ti) = 4540 kg/m³
ρ₂ = density of Aluminum (Al) = 2710 kg/m³
ρ₃ = density of Vanadium (V) = 6110 kg/m³
m₁ = mass fraction of Titanium (Ti) = 90% = 0.9
m₂ = mass fraction of Aluminum (Ti) = 6% = 0.06
m₁ = mass fraction of Vanadium (V) = 4% = 0.04
Therefore,
1/ρ = 0.9/(4540 kg/m³) + (0.06)/(2710 kg/m³) + (0.04)/(6110 kg/m³)
1/ρ = 1.9823 x 10⁺⁴ m³/kg + 0.2214 x 10⁻⁴ m³/kg + 0.0654 x 10⁻⁴ m³/kg
 1/ρ =  2.2691 x 10⁻⁴ m³/kg 
ρ = 1/(2.2691 x 10⁻⁴ m³/kg)
<u>ρ = 4407.03 kg/m³</u>
 
        
             
        
        
        
You have to figure out a way to write the two unknown abundances in terms of one variable. 
The total abundance is 1 (or 100%). So if you say the abundance for the first one is X then the abundance for the second one has to be 1-X (where X is the decimal of the percentage so say 0.8 for 80%). 
203(X) + 205(1-X) = 204.4 
Then you just solve for X to get the percentage for TI-203. 
And then solve for 1-X to get the percentage for TI-205. 
After that the higher percentage would be the most abundant.
203x + 205 - 205x = 204.4
-2x + 205 = 204.4
-2x = -0.6
x = 0.3
1-x = 0.7
Then the TI-205 would have the highest percentage and would be the most abundant.
        
             
        
        
        
Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
 
        
             
        
        
        
I believe the answer is compound B may have a lower molecular weight compared to compound A. 
At the same temperature, lighter particles of a compound have a higher average speeds than do heavier particles of another compound. Thus, particles of compound B are lighter than those of compound A and thus they have a higher average speed, hence evaporating faster compared to compound A.
        
                    
             
        
        
        
Answer:
nickel is the correct answer