You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.
For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.
6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2
For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.
13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2
As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.
In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!
Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M
Answer : The approximate relation Celsius = 1/2 Fahrenheit is a better approximation at higher temperatures
Explanation :
The formula for Celsius to Fahrenheit conversion is
At lower temperature the value that needs to be subtracted (32) is large enough as a result the approximation "celsius = 1/2 fahrenheit " does not seem valid.
For example, 50 F is 10°C.
This is almost 1/5 of Fahrenheit temperature.
But at higher temperatures , the value becomes insignificant and also the ratio 5/9 tend to be equal to 0.5.
For example, 2000 F is 1093°C
This is almost half of Fahrenheit temperature.
Therefore , the approximate relation Celsius = 1/2 Fahrenheit is a better approximation at higher temperatures
Answer:
0.67mol/Kg
Explanation:
The following were obtained from the question:
Mole of solute = 0.50mol
Mass of solvent = 750g = 750/1000 = 0.75Kg
Molality =?
Molality = mole of solute /mass of solvent
Molality = 0.5/0.75
Molality = 0.67mol/Kg
Answer:
The density of Lithium β is 0.5798 g/cm³
Explanation:
For a face centered cubic (FCC) structure, there are total number of 4 atoms in the unit cell.
we need to calculate the mass of these atoms because density is mass per unit volume.
Atomic mass of Lithium is 6.94 g/mol
Then we calculate the mass of four atoms;
⇒next, we estimate the volume of the unit cell in cubic centimeter
given the edge length or lattice constant a = 0.43nm
a = 0.43nm = 0.43 X 10⁻⁹ m = 0.43 X 10⁻⁹ X 10² cm = 4.3 X 10⁻⁸cm
Volume of the unit cell = a³ = (4.3 X 10⁻⁸cm)³ = 7.9507 X 10⁻²³ cm³
⇒Finally, we calculate the density of Lithium β
Density = mass/volume
Density = (4.6097 X 10⁻²³ g)/(7.9507 X 10⁻²³ cm³)
Density = 0.5798 g/cm³
