How many moles of H2SO4 are required to completely neutralize 0.10 moles of Ca(OH)2?

Chemistry

2 answers:

Kisachek [45]3 years ago

4 0

Answer:

0.10 mol of sulfuric acid will neutralize 0.10 mol of calcium hydroxide.

Explanation:

$H_2SO_4+Ca(OH)_2\rightarrow CaSO_4+2H_2O$

1 mol of sulfuric acid reacts or neutralizes with 1 mol of calcium hydroxide.

Then 0.10 mole of calcium hydroxide will be neutralized by:

$\frac{1}{1}\times 0.10 mol= 0.10 mol$ of sulfuric acid

0.10 mol of sulfuric acid will neutralize 0.10 mol of calcium hydroxide.

zalisa [80]3 years ago

3 0

Mole ratio:

H2SO4 + Ca(OH)2 = CaSO4 + 2 H2O

1 mole H2SO4 ----------- 1 mole Ca(OH)2
? moles H2SO4 --------- 0.10 moles Ca(OH)2

0.10 x 1 / 1

= 0.10 moles of H2SO4

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Draw the alkane formed when 4,5,5-trimethyl-1-hexyne is treated with two equivalents of hbr.

AveGali [126]

<h3>Answer:</h3>

The Alkane formed is 5,5-dibromo-2,2,3-trimethylhexane. as shown below in attached scheme (Green Color).

<h3>Explanation:</h3>

Alkynes like Alkenes undergo <em>Electrophillic Addition Reactions</em>. The reaction given is a two step reaction. In step 1, the Alkyne adds first equivalent of HBr obeying <em>Markovnikov's rule</em> (i.e. Bromine will add to carbon containing less number of hydrogen atoms) and forms <em>2-bromo-4,5,5-trimethylhex-1-ene</em>. In step 2, the alkene formed in first step (2-bromo-4,5,5-trimethylhex-1-ene) undergoes addition reaction with the second equivalent of HBr via Markovnikov's rule to produce <em>5,5-dibromo-2,2,3-trimethylhexane</em>.

The scheme is attached below, Blue color is assigned to starting Alkyne, Red color is assigned to intermediate Alkene and Green color is assigned to product Alkane respectively.

3 0

3 years ago

Read 2 more answers

For the generic equilibrium HA(aq) ⇌ H+(aq) + A−(aq), which of these statements is true? For the generic equilibrium , which of

timama [110]

<u>Answer:</u> The correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

<u>Explanation:</u>

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.

Equilibrium reaction of HA and KA follows the equation:

$HA\rightleftharpoons H^{+}(aq.)+A^{-}(aq.)$

$KA\rightleftharpoons K^+(aq.)+A^{-}(aq.)$

According to Le-Chateliers principle, if there is any change in the variables of the reaction, the equilibrium will shift in the direction in order to minimize the effect.

In the equilibrium reactions, $A^-$ ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of HA.

Thus, the addition of KA will shift the equilibrium in the left direction.

Equilibrium constant depends on the temperature of the system. It does not have any effect on any change of pH.

pH is defined as the negative logarithm of hydrogen ions present in the solution

If the solution has high hydrogen ion concentration, then the pH will be low.
If the solution has low hydrogen ion concentration, then the pH will be high.

As, the equilibrium is shifting in the left direction, that means concentration of $H^+$ ions are getting decreases. This will increase the pH of the solution.