I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research.All scientific theories <span>must be based on careful and rational examinations of the facts. I hope that this is the answer that you were looking for and it has actually come to your help.</span>
Answer:
We'll have 1 mol Al2O3 and 3 moles H2
Explanation:
Step 1: data given
Numer of moles of aluminium = 2 moles
Number of moles of H2O = 6 moles
Step 2: The balanced equation
2Al + 3H2O → Al2O3 + 3H2
Step 3: Calculate the limiting reactant
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
Aluminium is the limiting reactant. It will completely be consumed (2 moles).
H2O is in excess. There will react 3/2 * 2 = 3 moles
There will remain 6 - 3 = 3 moles
Step 4: Calculate moles products
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
For 2 moles Al we'll have 2/1 = 1 mol Al2O3
For 2 moles Al We'll have 3/2 * 2 = 3 moles H2
We'll have 1 mol Al2O3 and 3 moles H2
Answer:
a) The concentration of drug in the bottle is 9.8 mg/ml
b) 0.15 ml drug solution + 1.85 ml saline.
c) 4.9 × 10⁻⁵ mol/l
Explanation:
Hi there!
a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml
b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:
0.0210 mg/kg-body mass * 70 kg = 1.47 mg drug.
The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)
If 9.8 mg of the drug is contained in 1 ml of solution, then 1.47 mg drug will be present in (1.47 mg * 1 ml/ 9.8 mg) 0.15 ml.
To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline
c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.
Let´s convert it to molarity:
0.735 mg/ml * 1000 ml/l * 0.001 g/mg* 1 mol/ 15000 g = 4.9 × 10⁻⁵ mol/l
I assume what they are asking you? Sorry if that sound mean
First. let's write the reaction formula: HBr +LiOH ----> LiBr + H₂O
let's get the moles of LiOH first
moles= Molarity x Liters
moles= 0.253 M x 0.01673 Liter= 0.00423 moles LiOH
using the balanced equation, you can see that 1 mol LiOH is equal to 1 mol HBr. so:
0.00423 mol LiOH = 0.00423 mol HBr
now let's find the concentration
molarity= mol/ Liters
0.00423 mol/ 0.01000 Liters= 0.423 M
