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Ivenika [448]
3 years ago
6

A 1.000 mL sample of acetone, a common solvent sometimes used as a paint remover, was placed in a small bottle whose mass was kn

own to be 38.0015 g.
The following values were obtained when the acetone-filled bottle was weighed:
38.7798 g, 38.7795 g, and 38.7801 g.

How would you characterize the precision and accuracy of these measurements if the true mass of the acetone was 0.7791 g?
Chemistry
1 answer:
ra1l [238]3 years ago
7 0

Answer:

See explanation

Explanation:

Precision means the reputability of measurement and accuracy means how close a measurement is to the actual value. To get the measured mass of the acetone we need to substract the mass of the bottle from the measured mass of bottle and acetone.

The mass of the bottle is 38.0015g

⇒The mass of acetone in bottle 1= 0.7783g

⇒Mass of acetone in bottle 2= 0.7780g

⇒Mass of acetone in bottle3= 0.7786g

The measured value is near to each other. ⇒ the measurements are <u>precise</u>.

To check the <u>accuracy</u> we can compare the average value to the actual mass of the acetone.

Average of the acetone measurement is (0.7783+0.7780+0.7786)/3 = 0.7783g

The percentage of difference of the average measurement to the actual mass is = {(actual value-measured value)/actual value}x100%

={(0.7791-0.7783)/0.7791}x100%

=0.10%

So we can see the difference is very small ⇒ the measurement is accurate.

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How many oxygen atoms are there in 582 grams of caffeine (CH10N402, 194 g/mol)? Avogadro's Number: 1 mole = 6.02 x 1023 species
Sergeu [11.5K]

Answer:

3.61 * 10 ²⁴atoms.

Explanation:

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

w = 582 g

m = 194 g/mol

The number of moles can be calculated from the above formula , and substituting the respective values ,

n = w / m  = 582 g / 194 g/mol  =  3 mol

In the molecular formula of caffeine ,

<u>In 1 mole of caffeine their are - 2 moles of Oxygen. </u>

Therefore , in  3 moles of Caffeine there will be 6 moles of oxygen.

As well know ,

one mole of any substance contains 6.023*10²³ atoms,

Therefore , in  6 mol of oxygen =  6 * 6.023*10²³ atoms  = 3.61 * 10 ²⁴atoms.

8 0
3 years ago
The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C
artcher [175]

Explanation:

The given data is as follows.

         P_{1} = 100 mm Hg or \frac{100}{760}atm = 0.13157 atm

         T_{1} = 1080 ^{o}C = (1080 + 273) K = 1357 K

         T_{2} = 1220 ^{o}C = (1220 + 273) K = 1493 K

         P_{2} = 600 mm Hg or \frac{600}{760}atm = 0.7895 atm

          R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

                   log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}

            log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]

          log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]

                0.77815 = \frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K

              \Delta H_{vap} = 2.219 \times 10^{5} J/mol

                                   = 2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}

                                    = 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

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Answer:

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What is the name of the drug you were talking at the time of coming up with this theory please?

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