Answer:
94.1 %
Explanation:
We firstly determine the equation:
2H₂O + O₂ → 2H₂O₂
2 moles of water react to 1 mol of oxygen in order to produce 2 moles of oxygen peroxide.
We convert the mass of oxygen to moles:50 g . 1mol /32g = 1.56 mol
Certainly oxygen is the limiting reactant.
2 moles of water react to 1 mol of oxygen.
13 moles of water may react to 13/2 = 6.5 moles. (And we only have 1.56)
As we determine the limiting reactant we continue to the products:
1 mol of O₂ can produce 2 moles of H₂O₂
Then 1.56 moles of O₂ will produce (1.56 . 2) = 3.125 moles
We convert the moles to mass: 3.125 mol . 34 g/mol= 106.25 g
That's the 100% yield or it can be called theoretical yield.
Percent yield = (Yield produced / Theoretical yield) . 100
(100g / 106.25 g) . 100 = 94.1 %
Answer:
I don't know
Explanation:
pls I need a little bit answer
(4 mol H2O) x (112 kJ / 3 mol H2O) = 149 kJ
<span>(14.5 g HCl) / (36.4611 g HCl/mol) x (112 kJ / 3 mol HCl) = 14.9 kJ </span>
<span>(475 kJ) / (181 kJ / 2 mol HgO) x (216.5894 g HgO/mol) = 1137 g HgO </span>
<span>(179 kJ) / (181 kJ / 1 mol O2) x (31.99886 g O2/mol) = 31.6 g O2 </span>
<span>(145 kJ) / (112 kJ / 3 mol Cl2) x (70.9064 g Cl2/mol) = 275 g Cl2 </span>
<span>(14.5 g S2Cl2) / (135.0360 g S2Cl2/mol) x (112 kJ / 1 mol S2Cl2) = 12.0 kJ </span>
<span>CaCO3 + 2 NH3 → CaCN2 + 3 H2O; ∆H = –90.0 kJ </span>
<span>(798 kJ) / (90.0 kJ / 2 mol HN3) x (17.03056 g NH3/mol) = 302 g NH3 </span>
<span>(19.7 g H2O) / (18.01532 g H2O/mol) x (90.0 kJ / 3 mol H2O) = 32.8 kJ</span>
Answer:
its none above
Explanation:
food webs have arrows that show the flow of energy from one orhganism to the next
bar graphs have bars and numbers and diagrams have words and pictures so it D
The simplest unit factor that relates the number of hydrogen in sucrose to the number of oxygens is 1 O atom/ 2 H atoms.
Sucrose is a diasaccharide having the formula; C12H22O11.
This implies that there are twenty two atoms of hydrogen and eleven atoms of oxygen in a molecule of sucrose.
This gives a ratio of 2 hydrogen atoms to one oxygen atom. Therefore, the simplest unit factor that relates the number of hydrogen in sucrose to the number of oxygens is 1 O atom/ 2 H atoms.
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