Here, we should use combined gas law which can be derived from combined gas law, “PV=nRT”. Rearranging, we can get PV/T=nR. Then we can set the two states in the problem together to get
P1V1/T1 = P2V2/T2
Then just plug in and solve algebraically.
Hope this helps
Hello
the answer is 43.129310000000004
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Answer:

Explanation:
Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.
The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

The net equation is then:

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:
Actual anode half-equation:

Actual cathode half-equation:

Actual net reaction:

Explanation:
2.04 % hydrogen
32.65% sulphur
65.31% is oxygen
atomic ratio
hydrogen =2.04÷1=2.04
sulphur =32.65÷32=1.02
oxygen =65.31÷16=4.08
simplest ratio
hydrogen = 2.04÷1.02=2
sulphur =1.02÷1.02=1
oxygen =4.08÷1.02=4
empirical formula is H2SO4
Answer:
K = 10
Explanation:
Using Hess's law, it is possible to obtain the equilibrium constant, K, of a reaction using K of similar reactions. For example:
<em> If A ⇄ B K = X</em>
B ⇄ A K = 1/X
2A ⇄ 2B K = X².
Thus, if A(g) ⇄ 2B(g) K = 0.010
2B(g) ⇄ A(g) K = 1 / 0.010; K = 100
B(g) ⇄ A(g) K = √100 = 10
<h3>K = 10</h3>