2H2 + O2 -> 2H20 is the balanced equation for the reaction of Hydrogen with Oxygen to form water so If you have 32g of O2 this is a simple Dimensional analysis problem 32g O2 x 36.03056g H20/31.9988g O2 this way the O2 cancels out and you are left with just the H2O so your raw answer would be 36.0319112, then if your instructor requires a significant figure answer that would be to 2 significant figures the information you were given 32g O2, so as above 36g or Water are produced. Just a different way to view and solve the problem with the balanced equation so you can see the way everything relates to everything else. the molar masses of O2 and H2O are simply found my adding up 2 Oxygens 15.9994g x2 = 31.9988g and H2O = 2(1.00794) + 15.9994 = 18.01258 but you then have to multiply that by 2 because the reaction states you get 2 mols in the reaction so that is where I came up with the 36.03056g for the solution. Hope this helps. Dimensional Analysis is and can be one of the best ways to solve these problems, because not always are you going to be dealing with 1:2 ratios.
Answer:
1.36x10^10L
Explanation:
Step 1:
Determination of the mole of fluorine that contains 3.66x10^32 molecules. This is shown below:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10^23 molecules. This implies that 1 mole of fluorine also contains 6.02x10^23 molecules.
Now if 1 mole of fluorine contains 6.02x10^23 molecules,
Therefore, Xmol of fluorine will contain 3.66x10^32 molecules i.e
Xmol of fluorine = 3.66x10^32/6.02x10^23
Xmol of fluorine = 6.08x10^8 moles
Step 2:
Determination of the volume occupied by 6.08x10^8 moles of fluorine.
1 mole of any gas occupy 22.4L at stp. This means that 1 mole of fluorine also occupy 22.4L at stp.
Now if 1 mole of fluorine occupies 22.4L at stp,
Then 6.08x10^8 moles of fluorine will occupy = 6.08x10^8 x 22.4 = 1.36x10^10L
Answer:
None of these
Explanation:
For a reaction;
aA + bB ------>cC + dD
The equilibrium constant K is given as;
K = [C]^c [D]^d/[A]^a [B]^b
The equilibrium constant neither depends on the concentrations of the reactants nor on that of the products.
Let us recall that at equilibrium, the concentrations of reactants and products remain largely constant. This implies that, concentration of species do not appreciably change at equilibrium because the rates of forward and reverse reactions are equal.
Hence, the equilibrium constant neither depends on the initial/final concentrations of the reactants nor on the initial/final concentrations of the products.
The only thing you need to know in order to calculate <span>the gravimetric factor is the formula :
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I am pretty sure that you will find it helpful! Regards.
That is, by definition, true.
