Answer:
2 moles of Sn are produced when 4 moles of H2(g) are consumed completely
Explanation:
to determine the number of moles of sn (l) produced when 4.0 moles of H2 (g) is consumed completely.
First, find the number of moles of H2 consumed by taking this as limiting reagent.
Then find the moles of Sn (l) taking into account the stoichiometric relationship between H2(g) and Sn(l). 2:1
(s) + 2
(g) ⇒ Sn(l) + 2
(g)
∴2 moles of Sn are produced when 4 moles of H2(g) are consumed completely.
To determine the absolute pressure of this gas, all you need to do is to add the value of atmospheric pressure and the value of gage pressure.
Atmospheric pressure is equivalent to 100 kPa.
Gage pressure is 276 kPa.
Then, we add both values.
N = 100 kPa + 276 kPa
N = 376 kPa
The absolute pressure of this gas is 376 kPa.
Hope this helps :)
What question? Lhh this is hilarious.
Answer:
it is 50
if I m right pls mark me brainliest
Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
