According to the equation you have 1 mole of C2H4 and 3 moles of O2.
1 • (22.4L / 270L) = 3 • (22.4L / x)
1/270L = 3/x
x = 3(270) / 1
x = 810 L
810 Liters of oxygen will react with 270 liters of ethene (C2H4) at STP
I don’t get the question. is there a worksheet?
Answer:
No matter how many times you cut it, its chemical properties won't change and it'll still be paper.
Explanation:
Answer:
The concentration of the CaBr2 solution is 96 µmol/L
Explanation:
<u>Step 1:</u> Data given
Moles of Calciumbromide (CaBr2) = 4.81 µmol
Volume of the flask = 50.0 mL = 0.05 L
<u>Step 2:</u> Calculate the concentration of Calciumbromide
Concentration CaBr2 = moles CaBr2 / volume
Concentration CaBr2 = 4.81 µmol / 0.05 L
Concentration CaBr2 = 96.2 µmol /L = 96.2 µM
The concentration of the CaBr2 solution is 96 µmol/L
Answer:
6Br⁻ + XeO₃ + 6H⁺ → 3Br₂ + Xe + 3H₂O
Explanation:
First, we need to write the half-reactions:
2Br⁻ → Br₂ + 2e⁻ Oxidation -Balanced yet-
XeO₃ → Xe Reduction
To balance the reduction in acidic aqueous solution we need to add waters in the other side of the reaction as oxygens are present:
XeO₃ → Xe + 3H₂O
And H⁺ as hydrogens from water we have:
XeO₃ + 6H⁺ → Xe + 3H₂O
To balance the charge:
<h3>XeO₃ + 6H⁺ + 6e⁻ → Xe + 3H₂O Reduction -Balanced-</h3><h3 />
To cancel out the electrons of both half-reaction we need to multiply oxidation 3 times:
6Br⁻ → 3Br₂ + 6e⁻
XeO₃ + 6H⁺ + 6e⁻ → Xe + 3H₂O
And the balanced reaction in acidic aqueous solution is the sum of both half-reactions:
<h3>6Br⁻ + XeO₃ + 6H⁺ → 3Br₂ + Xe + 3H₂O </h3>
