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2 years ago

Precision Problems:

Chemistry

1 answer:

Lorico [155]2 years ago

4 0

Answer:

$Precision = 2.70 \± 0.1\ cm$

Explanation:

Given

The data in the table

Required

Follow the steps appended to the question;

Step 1: Calculate the Mean or Average

Mean = Summation of lengths divided by number of teams;

$Mean = \frac{2.65 + 2.75 + 2.80 + 2.77 + 2.60 + 2.65 + 2.68}{7}\ cm$

$Mean = \frac{18.9}{7}\ cm$

$Mean = 2.70\ cm$

Step 2: Get The Range

$Range = Highest - Lowest$

$Range = 2.80cm - 2.60cm$

$Range = 0.2\ cm$

Step 3: Divide Range by 2

$Approximate\ Range = \frac{1}{2}Range$

$Approximate\ Range = \frac{1}{2} * 0.2\ cm$

$Approximate\ Range = 0.1\ cm$

Step 4: Determine the Precision

$Precision = Average \± Approximate\ Range$

Substitute 2.70 for Average and 0.1 for Approximate Range

$Precision = 2.70 \± 0.1\ cm$

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The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k

Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

$ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

We have given in the question

$P_1=102\ mmHg$

$T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol$

And $R$ is the Universal Gas Constant.

$R=0.008 314 kJ/Kmol$

$ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165$