hey there!:
2HgO (s) => 2Hg (l) + O2 (g)
2 moles of HgO decompose to form 2 moles of Hg and 1 mole of O2 according to the reaction mentioned in the question.
So 4.00 moles of HgO must give 4 moles of Hg and 2 moles of O2 theoretically.
603 g of Hg = 603 / 200.6 = 3 moles
Percent yield = ( actual yield / theoretical yield) * 100
= ( 3/4) * 100
= 75 %
Hope this helps!
All you can conclude is that something must be burning with an orange flame.
Actually, the "something" that must be burning is the hydrogen that is produced when the sodium reacts with the water:
2Na + 2H₂O → 2NaOH + H₂ + heat
So much heat is produced that the hydrogen catches fire and some of the sodium evaporates into the flame.
The electrons in the sodium atoms get "excited" in the flame. When they drop back to a lower energy level, they emit energy in the form of an orange-yellow light.
Explanation:
animals; move
I THINK SO SORRY IF I AM WRONG
Answer:
a. H2S(g)/t = 1.48 mol/s
CS2(g)/t = 0.740mol/s
H2(g)/t = 2.96mol/s
b.
Ptot /t = 981torr/min
Explanation:
a. Based on the reaction:
CH4(g) + 2 H2S(g) → CS2(g) + 4 H2(g)
<em>1 mole of CH4 reacts with 2 moles of H2S producing 1 mole of CS2 and 4 moles of 4H2</em>
<em />
If CH4 decreases at the rate of 0.740mol/s, H2S decreases twice faster, that is 0.740mol/s = 1.48 mol/s
CS2 is produced with the same rate of CH4 because 1 mole of CH4 produce 1 mole of CS2 = 0.740mol/s
The H2 is produced four times faster than CH4 is decreased, that is:
0.740mol/s * 4 = 2.96mol/s
b. With the reaction:
2 NH3(g) → N2(g) + 3 H2(g)
2 moles of ammonia are consumed whereas 1 mole of N2 and 3 moles of H2 are produced.
That means 2 moles of gas are consumed and 4 moles of gas are produced.
If the NH3 decreases at a rate of 327torr/min, the gases are produced in a rate twice faster. That is 327torr/min*2 =
654torr/min
The rate of change of the total pressure is rate of reactants + rate of products:
654torr/min + 327torr/min =
981torr/min
Yes, the crystals will have more time to form and therefore are more likely to be larger compared to one that is cooled quickly.
