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sergejj [24]
2 years ago
11

In the titration of a solution of Sodium hydroxide, an acid titrant was prepared by diluting 125.0 mL of 10.00 mol/L Nitric acid

into enough distilled water to make 500.0 mL of solution. 30.0 mL of the base was measured, and in the titration, 16.74 mL of the acid titrant was needed to neutralize. Determine the concentration of the Sodium hydroxide.
Chemistry
1 answer:
harina [27]2 years ago
6 0

Answer:

1.395M NaOH

Explanation:

Sodium hydroxide, NaOH, reacts with nitric acid, HNO3, as follows:

NaOH + HNO3 → NaNO3 + H2O

<em>Where 1mol of NaOH reacts with 1mol of HNO3</em>

To solve this question we must find the concentration of the titrant. With the concentration and the needed acid we can find the moles of HNO3 added = moles NaOH in the solution. With the moles of NaOH and its volume we can find its concentration as follows:

<em>HNO3 concentration:</em>

10.00mol/L HNO3 * (125.0mL/500.0mL) = 2.500M HNO3

<em>Moles HNO3 = Moles NaOH:</em>

16.74mL = 0.01674L*(2.500mol/L) = 0.04185 moles HNO3 = Moles NaOH

<em>Concentration NaOH:</em>

0.04185 moles / 0.0300L =

1.395M NaOH

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Both 1,2−dihydronaphthalene and 1,4−dihydronaphthalene may be selectively hydrogenated to 1,2,3,4−tetrahydronaphthalene. One of
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Answer:

1,4-dihydro = 113 kJ·mol⁻¹

1,2-dihydro = 101 kJ·mol⁻¹

Explanation:

In 1,4-dihydronaphthalene, the 2,3-double bond is isolated from the benzene ring.

In 1,2-dihydronaphthalene, the 3,4-double bond is conjugated with the benzene ring.

Thus, 1,2-dihydronaphthalene is partially stabilized by resonance interactions between the ring and the double bond (think, styrene).

1,2-Dihydronaphthalene is at a lower energy level because of this stabilization.

The heat of hydrogenation of 1,2-dihydronaphthalene is therefore less than that of the 1,4-isomer when each is hydrogenated to the common product, 1,2,3,4-tetrahydronaphthalene.

8 0
3 years ago
A stock solution has a concentration of 1.5 M NaCl and is diluted to a 0.80 M solution with a volume of 0.10 L. What volume of t
ioda

Answer:

0.053 L  is the volume of concentrated solution that was used

Explanation:

Let's determine the answer of this, by rules of three.

There is also a dilution formula.

Molarity is a sort of concentration that indicates the moles of solute in 1L of solution.

In 1 L of concentrated solution, there are 1.5 moles of NaCl

In 1 L of diluted solution, there are 0.80 moles.

The volume for the diluted solution is 0.10L

The rule of three will be:

1L of solution has 0.80 moles of solute

Then, 0.10L of solution must have (0.1 . 0.8)/1 = 0.08 moles

This moles came from the concentrated solution, and we know that in 1L of this solution we have 1.5 moles. Therefore the rule of three will be:

1.5 moles are in 1L of solution

0.08 moles were in (0.08 . 1L / 1.5) = 0.053 L (This is the volume of concentrated solution that was used)

Dilution formula is: M conc . Vol conc = M diluted . Vol diluted

1.5 M . Vol conc = 0.80 M . 0.10L

Vol conc = 0.80 M . 0.10L / 1.5M = 0.053L

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3 years ago
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A patient needs to be given exactly 1000 ml of a 5.0% (w/v) intravenous glucose solution. the stock solution is 55% (w/v). how m
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The patient needs 1000 ml of 5% (w/v) glucose solution
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