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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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Answer:

Approximately 3.81\; \rm m.

Explanation:

Look up the density \rho of carbon tetrachloride, \rm CCl_4, and glycerol:

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Let g denote the gravitational field strength. (Typically g \approx 9.81\; \rm N \cdot kg^{-1} near the surface of the earth.) For a column of liquid with a height of h, if the density of the liquid is \rho, the pressure at the bottom of the column would be:

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The pressure at the bottom of this carbon tetrachloride column would be:

\begin{aligned} P &= \rho \cdot g \cdot h \\ & \approx 1.59\times 10^{3} \; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1} \times 3.02 \; \rm m \approx 4.71 \times 10^{4} \; \rm N \cdot m^{-2} \end{aligned}.

Rearrange the equation P = \rho\cdot g \cdot h for h:

\displaystyle h = \frac{P}{\rho \cdot g}.

Apply this equation to calculate the height of the liquid glycerol column:

\begin{aligned}h &= \frac{P}{\rho \cdot g} \\ &\approx \frac{4.71 \times 10^{4}\; \rm N \cdot m^{-2}}{1.26 \times 10^{3}\; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1}} \approx 3.81\; \rm m\end{aligned}.

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Hi there!
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Answer:

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