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2 years ago

10

A thermometer that was used in a calorimeter was consistently 2.7 °c too high. how would this affect the calculations of δh for

the reactions conducted in the calorimeter?

1 answer:

vodomira [7]2 years ago

7 0

The calorimetry experiment is usually done to determine the heat capacity of the sample. The working equation would be:

Q = mCpΔT
Q is the energy
m is the mass of sample
Cp is the heat capacity
ΔT is the temperature

So, if the thermometer is too high, then that would affect ΔT, which would make it greater. Consequently, you would calculate a much lesser heat capacity of the sample compared to the theoretical value.

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Which one of these statements about carbon tetrachloride fire extinguishers is true?

Fofino [41]

Answer;

-Prohibited from use by OSHA

Explanation;

-Carbon tetrachloride is a clear and colorless liquid with an odor similar to that of chloroform.It is employed as a chemical reagent for a number of purposes, as a raw material in chemical manufacture and is, used very widely as a reagent to extinguish fires.

-It volatizes very easily, is a non-conductor of electricity and, as noted above, freezes at very low temperatures.

-It is effective, due to the blanketing effect of its vaporized fumes, on many Class B or volatile, flammable liquid fires and because it will mix with certain volatile, flammable liquids and form a non-burning combination.

5 0

3 years ago

The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.

olchik [2.2K]

Answer:

Approximately $0.291\; \rm M$ (rounded to two significant figures.)

Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

5 0

3 years ago

In the summer of 1859, Edwin Drake became the first person to strike oil in the United States while drilling in Titusville,_____

Nata [24]

C.Pennsylvania Titusville is in the state of Pennsylvania

7 0

3 years ago

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An unknown amount of acid can often be determined by adding an excess of base and then back-titrating the excess. A 0.3471−g sam

MAXImum [283]

Explanation:

The given data is as follows.

Mass of mixture = 0.3471 g

As the mixture contains oxalic acid and benzoic acid. So, oxalic acid will have two protons and benzoic acid has one proton.

This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.

Hence, moles of NaOH in 97 ml = $\frac{0.1090 \times 97}{1000}$

= $10.573 \times 10^{-3} mol$

Moles of HCl in 21.00 ml = $\frac{0.2060 \times 21}{1000}$

= $4.326 \times 10^{-3}$ mol

Therefore, total moles of NaOH that reacted are as follows.

$10.573 \times 10^{-3} mol$ - $4.326 \times 10^{-3} mol$

= $6.247 \times 10^{-3}$ mol

So, total 3 mole of NaOH will react with 1 mole of mixture. Therefore, number of moles of NaOH reacted with benzoic acid is as follows.

$\frac{6.247 \times 10^{-3}}{3}$

= $2.082 \times 10^{-3}$ mol

Since, molar mass of NaOH is 40 g/mol. Therefore, calculate the mass of NaOH as follows.

$2.082 \times 10^{-3} mol \times 40 g/mol$

= $83.293 \times 10^{-3}$ g

= 0.0832 g

Whereas molar mass of benzoic acid is 122 g/mol.

Therefore, 40 g NaOH = 122 g benzoic acid

So, 0.0832 g NaOH = $\frac{122 g}{40 g} \times 0.0832 g$

= 0.253 g

Hence, calculate the % mass of benzoic acid as follows.

$\frac{0.253 g}{0.3471 g} \times 100$

= 73.10%

Thus, we can conclude that mass % of benzoic acid is 73.10%.

4 0

3 years ago

Which statement best describes the properties of ionic compounds? They are brittle, solid at room temperature, and have a high m

djverab [1.8K]

They are good conductors of heat and electricity.

They are solid at room tempature

They have a high melting point

6 0

3 years ago

Read 2 more answers