To solve this we use the
equation,
M1V1 = M2V2
where M1 is the concentration of the stock solution, V1 is the
volume of the stock solution, M2 is the concentration of the new solution and
V2 is its volume.
2.5 M x V1 = 1.0 M x .250 L
<span>V1 = 0.10 L or 100 mL of the 2.5 M HCl solution is needed
Hope this helps.</span>
The amount of Al2O3 in moles= 1.11 moles while in grams = 113.22 grams
<em><u>calculation</u></em>
2 Al + Fe2O3 → 2Fe + Al2O3
step 1: find the moles of Al by use of <u><em>moles= mass/molar mass </em></u>formula
= 60.0/27= 2.22 moles
Step 2: use the mole ratio to determine the moles of Al2O3.
The mole ratio of Al : Al2O3 is 2: 1 therefore the moles of Al2O3= 2.22/2=1.11 moles
Step 3: finds the mass of Al2O3 by us of <u><em>mass= moles x molar mass</em></u><em> </em>formula.
The molar mass of Al2O3 = (2x27) +( 16 x3) = 102 g/mol
mass is therefore= 102 g/mol x 1.11= 113.22 grams
<u><em>Answer: Chemical reaction, a process in which one or more substances, the reactants, are converted to one or more different substances, the products.</em></u>
Explanation:
The answer is 60 kpa, also did you try look it up because that is what i did but i did not copy and paste it.
hope this helped have a good day
Answer : The excess reactant in the combustion of methane in opem atmosphere is 
molecule.
Solution : Given,
Mass of methane = 23 g
Molar mass of methane = 16.04 g/mole
The Net balanced chemical reaction for combustion of methane is,
First we have to calculate the moles of methane.
= 
= 1.434 moles
From the above chemical reaction, we conclude that
1 mole of methane react with the 2 moles of oxygen
and 1.434 moles of methane react to give 
moles of oxygen
The Moles of oxygen = 2.868 moles
Now we conclude that the moles of oxygen are more than the moles of methane.
Therefore, the excess reactant in the combustion of methane in open atmosphere is molecule.
