The pressure of the nitrogen gas produced is determined as 44.77 atm.
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What is the pressure of the Nitrogen gas?</h3>
The pressure of the nitrogen gas is determined from ideal gas equation, as shown below;
PV = nRT
P = nRT/V
where;
- n is number of moles = 2 moles
- R is ideal gas constant = 0.08205 L.atm/mol.K
- T is temperature = 68⁰C = 68 + 273 = 341 K
- V is volume = 1.25 L
P = (2 x 0.08205 x 341)/(1.25)
P = 44.77 atm.
Learn more about pressure here: brainly.com/question/25736513
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Answer:
Balanced reaction:
3 H2 (g) + N2 (g) → 2 NH3 (g)
Use stoichiometry to convert g of H2 to g of NH3. The process would be:
g H2 → mol H2 → mol NH3 → g NH3
12.0 g H2 x (1 mol H2 / 2.02 g H2) x (2 mol NH3 / 3 mol H2) x (17.03 g NH3 / 1 mol NH3) = 67.4 g NH3
Explanation: See above
Hope this helps, friend.
Answer: acid
Explanation: it dissolves in water to produce hydronium ion as the only positive ion
The answer would be B.
U-238 has a n to p ration of 1.6:1. 146 neutrons and 92 protons.
It is actually the most commonly used isotope is reactors.
C-14 is also a radioactive isotope with 8 neutrons and 6 protons.
The usual and ideal n to p ratio is 1:1 such as C-12 or Mg-24
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:
