The balanced equation for the neutralisation reaction is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
the number of moles of NaOH reacted - 0.126 mol/L x 0.0173 L = 0.00218 mol
if 2 mol of NaOH reacts with 1 mol of H₂SO₄
then 0.00218 mol of NaOH reacts with - 0.00218 / 2 = 0.00109 mol of H₂SO₄
molarity is the number of moles of solute in 1 L solution
therefore if 25 mL contains - 0.00109 mol
then 1000 mL contains - 0.00109 mol / 25 mL x 1000 mL = 0.0436 mol/L
therefore molarity of H₂SO₄ is 0.0436 M
Is ionic equation . nitric acid + solid aluminium = hydroxide.
As per the question, the mass of the nitrogen gas m = 22.25 gram.
The latent heat of vaporization of nitrogen = 199.0 j/g
As per the question, the nitrogen gas will condense. During condensation, the nitrogen gas will lose or release heat equal to its latent heat.
Hence, the heat released by nitrogen gas Q = ml = 22.25 × 199.0 J = 4427.75 J.
Hence, the amount of heat released will be 4427.75 J.
<h3>How can you figure out how much heat is in each gram?</h3>
The formula: can be utilized to determine energy. Q = mc ∆T. In the equation, Q stands for energy expressed in joules or calories, m for mass expressed in grams, c for specific heat, and T for temperature change, which is the difference between the final temperature and the initial temperature. Water has a specific heat of 1 calorie/gram °C.
Learn more about energy here:
brainly.com/question/1932868
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<span>subshells have two reasons of stability-
</span><span> 1) half filled and full filled subshells lead to symmetry as symmetry always increases stability in the nature.
</span>2)<span> in half filled and full filled subshells there are maximum no Of exchanges so exchanges releases energy so it increase stability.. Exchange of e-s also can b understood as delocalization of e-s which is similar like resonance and resonance always increases stability</span>
Answer:
See attached picture.
Explanation:
Hello,
In this case, on the attached picture you will find the required line structures for the cis and trans configurations of the given compound (2-pentene). Take into account for the cis that the adjacent carbons to those having the double bond remain in the same plane, whereas for the trans one, the adjacent carbons remain in a different plane.
Regards-
