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Aliun [14]
3 years ago
5

What is the change of e-

Chemistry
1 answer:
Afina-wow [57]3 years ago
6 0
E- is a negative electron
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Answer:

A?

Explanation:

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What value must be known if the accuracy of an experiment is to be determined​
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To evaluate the accuracy of a measurement, the measured value must be compared to the correct value. To evaluate the precision of a measurement, you must compare the values of two or more repeated measurements.
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3 years ago
Boyle's law only works when a gas is kept at a constant temperature. Experimentally this is very tricky as changes in pressure o
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The heat that creates this temperature change coming from change in the internal energy of the system as per as first law of thermodynamics.

<h3>What is Boyle's law ?</h3>

A law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.

As we know, Boyle's law only works when the gas is kept at a constant temperature

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When volume of gases decreased, it means work done has occurred on the system, so the work done is used for raising internal energy of the gas and the other is released as the thermal energy.

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According to 1st law of thermodynamics,

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2 years ago
The conversation of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-⁴s-¹. a) i
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Answer: a)  The concentration after 8.8min is 0.17 M

b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) concentration after 8.8 min:

8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}

\log\frac{0.25}{a-x}=0.15

\frac{0.25}{a-x}=1.41

(a-x)=0.17M

b) for concentration to decrease from 0.25M to 0.15M

t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20

t=687s

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3 years ago
Using your knowledge of chemistry and the information in Reference Table H, which statement concerning propanone and water at 50
stiv31 [10]
The answer is (2) higher vapor pressure and weaker intermolecular forces. Propanone has a lower boiling point, so it is more volatile than water. Propanone's vapor pressure is, therefore, higher than that of water at 50 degrees Celsius. Propanone is more volatile due to the fact that the intermolecular forces that hold its molecuels together are not as strong as those that hold together molecules of water. Since the IMFs are weaker, it takes less thermal energy to break individual molecules free of each other.
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