Answer is (4) - Pb.
According to the reactivity series of elements
- the elements which are above the hydrogen are more reactive than hydrogen.
- the elements which are below the hydrogen are less reactive than hydrogen.
Among the given choices, only Pb is placed above the hydrogen in the reactivity series and rest are below the hydrogen.
Hence, Pb is more active than hydrogen.
Answer:
6.57 L
Explanation:
First, calculate the <em>moles of hydrogen</em> produced, then use the Ideal Gas Law to calculate the <em>volume of hydrogen</em>.
Step 1. Write the <em>chemical equation</em>.
: 22.99
2Na + H₂O ⟶ 2NaOH + H₂
Step 1. Convert <em>grams of Na</em> to <em>moles of Na</em>
Step 2. Use the molar ratio of H₂:Na to convert <em>moles of Na</em> to <em>moles of H₂</em>.
Step 3. Use the Ideal Gas Law to calculate the <em>volume of hydrogen</em>.
<em>pV = nRT</em>
Answer:
La Nina
Explanation:
La Nina -
The word La Nina means " the little girl " , in spanish language .
La Nina , is an ocean - atmospheric process , which occurs in the eastern equatorial Pacific Ocean .
It is exactly opposite of El Nino , which means " the little boy " .
In the phase of La Nina , the temperature of the eastern equatorial Pacific Ocean becomes lower than the usual temperature by around 3°C to 5°C .
This phase of La Nina is for around 5 months in an year .
Hence , from the given statement of the question ,
The correct term is La Nina .
Answer:
The pressure of N₂O₄ in the reaction vessel after the reaction is 290 mmHg
Explanation:
Nitrogen gas reacts with oxygen gas to form dinitrogen tetroxide.
N₂ (g) + 2O₂ (g) → N₂O₄ (g)
Therefore since by Avogadro's law equal volumes of all gases contain equal numbers of molecules, there fore as the gases are within the same vessel, thier partial pressure is equivalent to their concentration
from the reaction, 1 mole of N₂ react with 2 moles of O₂ to produce 1 mole of N₂O₄
Thus
1 mmHg of N₂ react with 2 mmHg of O₂ to produce 1 mmHg of N₂O₄
337 mmHg N₂ ×(1 mmHg of N₂O₄/ 1 mmHg of N₂) = 337 mmHg N₂O₄
580 mmHg O₂ ×(1 mmHg of N₂O₄/ 2 mmHg of O₂) = 290 mmHg N₂O₄
As seen from the above calculation, the limting reactant is oxygen and the partial pressure of N₂O₄ = 290 mmHg