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ella [17]
2 years ago
15

the student performed a second trial and accidentally added more sodium hydroxide to the flask than was needed to reach the end

point but recorded the final volume. would this error increase, decrease or have no effect on the calculated acid concentration for the second trial?
Chemistry
1 answer:
Harrizon [31]2 years ago
4 0

A greater volume of acid will be needed to neutralize the sodium hydroxide thereby making the calculated concentration of acid for the second trial to decrease.

Titration is usually a way of determining the concentration of an unknown solution by accurately measuring the volume of a known solution that reacts completely with the unknown solution.

The point when reaction is completed is indicated by the end point of the reaction. If more base (sodium hydroxide) is added to the flask, then more volume of acid will be needed to neutralize it thereby making the calculated concentration of acid for the second trial to decrease.

Learn more: brainly.com/question/1527403

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how many molecules of sulfuric acid are in a spherical raindrop of diameter 6.0 mm if the acid rain has a concentration of 4.4 *
Vitek1552 [10]

Answer:

The number of moles =

Moles=4.97\times 10^{-8}

The number of molecules =

Molecules = 2.99\times 10^{16}

Explanation:

Volume of the sphere is given by :

V=\frac{4}{3}\pi r^{3}

here, r = radius of the sphere

radius=\frac{diameter}{2}

radius=\frac{6.0}{2}

Radius = 3 mm

r = 3 mm

1 mm = 0.01 dm (1 millimeter = 0.001 decimeter)

3 mm = 3 x 0.01 dm = 0.03 dm

r = 0.03 dm

<em>("volume must be in dm^3 , this is the reason radius is changed into dm"</em>

<em>"this is done because 1 dm^3 = 1 liter and concentration is always measured in liters")</em>

V=\frac{4}{3}\pi 0.03^{3}

V=\frac{4}{3}\pi 2.7\times 10^{-5}

V=1.13\times 10^{-4}dm^{3}

V=1.13\times 10^{-4}L   (1 L = 1 dm3)

Now, concentration "C"=

C=4.4\times 10^{-4}moles/liter  

The concentration is given by the formula :

C=\frac{moles}{Volume(L)}

This is also written as,

Moles = C\times Volume

Moles=1.13\times 10^{-4}\times 4.4\times 10^{-4}

Moles=4.97\times 10^{-8}moles

One mole of the substance contain "Na"(= Avogadro number of molecules)

So, "n"  mole of substance contain =( n x Na )

N_{a}=6.022\times 10^{23}

Molecules =

Molecule=4.97\times 10^{-8}\times 6.022\times 10^{23}

Molecules = 2.99\times 10^{16} molecules

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