Answer:
Aluminium atoms = 4.13 *10^22 aluminium atoms
The correct answer is E
Explanation:
Step 1: Data given
Mass of Al2O3 = 3.50 grams
Molar mass of Al2O3 = 101.96 g/mol
Number of Avogadro = 6.022 * 10^23 /mol
Step 2: Calculate moles Al2O3
Moles Al2O3 = mass Al2O3 / molar mass Al2O3
Moles Al2O3 = 3.50 grams / 101.96 g/mol
Moles Al2O3 = 0.0343 moles
Step 3: Calculate moles Aluminium
In 1 mol Al2O3 we have 2 moles Al
in 0.0343 moles Al2O3 we have 2*0.0343 = 0.0686 moles Al
Step 4: Calculate aluminium atoms
Aluminium atoms = moles aluminium * Number of Avogadro
Aluminium atoms = 0.0686 * 6.022 * 10^23
Aluminium atoms = 4.13 *10^22 aluminium atoms
The correct answer is E
Using the combined gas law, where PV/T = constant, we first solve for PV/T for the initial conditions: (4.50 atm)(36.0 mL)/(10.0 + 273.15 K) = 0.57213.
Remember to use absolute temperature.
For the final conditions: (3.50 atm)(85.0 mL)/T = 297.5/T
Since these must equal, 0.57213 = 297.5/T
T = 519.98 K
Subtracting 273.15 gives 246.83 degC.
Answer:
Drug calculation
If we have 45g of clobetasol = 0.05%w/w
Then what mass in g of clobetasol is in 0.03%w/w = 45 x 0.03/0.05 =27g
It means that 27g of clobetasol must be added to change the drug strength to 0.03% w/w
Answer:
Explanation:
We have in this question the equilibrium
X ( g ) + Y ( g ) ⇆ Z ( g )
With the equilibrium contant Kp = pZ/(pX x pY)
The moment we change the concentration of Y, we are changing effectively the partial pressure of Y since pressure and concentration are directly proportional
pV = nRT ⇒ p = nRT/V and n/V is molarity.
Therefore we can calculate the reaction quotient Q
Qp = pZ/(pX x pY) = 1/ 1 x 0.5 atm = 2
Since Qp is greater than Kp the system proceeds from right to left.
We could also arrive to the same conclusion by applying LeChatelier´s principle which states that any disturbance in the equilibrium, the system will react in such a way to counteract the change to restore the equilibrium. Therefore, by having reduced the pressure of Y the system will react favoring the reactants side increasing some of the y pressure until restoring the equilibrium Kp = 1.
