Water has a vapor pressure of 23.8 torr at 298 K. What is the vapor pressure of water at 333 K? For water AHvap = 40.66 kJ/mol. Appropriate value of R is 0.008314 kJ/mol K. (Enter your answer to three significant figures.) Vapor pressure = torr

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
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Summary
A useful method to express a number is in logarithmic format. A logarithm is the exponent that a base must be raised to produce a given number.
Mathematically, if base = y then logbase (y) = x. For example we know that 10² 100 then the log10 (100) = 2. Likewise since 10-² = 0.01 then
log10 (0.01) = −2. Commonly for a base of 10 the logarithm symbol log is used. Thus for 10¹ = 10,000 the log₁0 (10,000) = 4.
=
Logarithms are not confined to base 10. For 3²
9 then log3 (9) = 2. A commonly used scientific base is Euler's number, 2.71828... Logarithms based upon
Euler's number are called natural logarithms having the symbol In. The antilogarithm of an In employs the exponential function e as found on every scientific
= 2.
0.693
calculator. For example the antilogarithm of 0.693 is e
Math Example
Enter the value of ln 130..
(Enter your answer to three significant figures.)
4.86
Enter the value of e³.0.
(Enter your answer to two significant figures.)
20
=
Chemistry Example
Water has a vapor pressure of 23.8 torr at 298 K. What is the vapor pressure of water at 333 K? For water Hvap = 40.66 kJ/mol. Appropriate value of Ris
0.008314 kJ/mol K.
(Enter your answer to three significant figures.)
Vapor pressure =
torr
Temperature: =
=
Water has a vapor pressure of 23.8 torr at 298 K. At what Kelvin temperature will water have a vapor pressure of 30.0 torr? For water Hvap :
. Appropriate value of R is 0.008314 kJ/mol K.
(Enter your answer to three significant figures.)
K
40.66 kJ/mol
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Transcribed Image Text:Summary A useful method to express a number is in logarithmic format. A logarithm is the exponent that a base must be raised to produce a given number. Mathematically, if base = y then logbase (y) = x. For example we know that 10² 100 then the log10 (100) = 2. Likewise since 10-² = 0.01 then log10 (0.01) = −2. Commonly for a base of 10 the logarithm symbol log is used. Thus for 10¹ = 10,000 the log₁0 (10,000) = 4. = Logarithms are not confined to base 10. For 3² 9 then log3 (9) = 2. A commonly used scientific base is Euler's number, 2.71828... Logarithms based upon Euler's number are called natural logarithms having the symbol In. The antilogarithm of an In employs the exponential function e as found on every scientific = 2. 0.693 calculator. For example the antilogarithm of 0.693 is e Math Example Enter the value of ln 130.. (Enter your answer to three significant figures.) 4.86 Enter the value of e³.0. (Enter your answer to two significant figures.) 20 = Chemistry Example Water has a vapor pressure of 23.8 torr at 298 K. What is the vapor pressure of water at 333 K? For water Hvap = 40.66 kJ/mol. Appropriate value of Ris 0.008314 kJ/mol K. (Enter your answer to three significant figures.) Vapor pressure = torr Temperature: = = Water has a vapor pressure of 23.8 torr at 298 K. At what Kelvin temperature will water have a vapor pressure of 30.0 torr? For water Hvap : . Appropriate value of R is 0.008314 kJ/mol K. (Enter your answer to three significant figures.) K 40.66 kJ/mol Previous Next
The reaction
2 NO(g) + O2(g) → 2 NO2 (g)
was studied, and the following data were obtained where
Rate
Δ[02]
At
[NO]。
[0₂]0
(molecules/cm³) (molecules/cm³)
1.00 × 10¹8
3.00 × 1018
2.50 × 1018
Submit Answer
1.00 × 10¹8
1.00 × 1018
2.50 × 1018
Initial Rate
What would be the initial rate for an experiment where [NO]。 = 8.18 × 10¹8 molecules/cm³ and [0₂]0 = 3.57 × 10¹8 molecules/cm³?
Rate =
molecules/cm
cm³ • S
Try Another Version
(molecules/cm³.s)
2.00 × 10¹6
1.80 × 10¹7
3.13 x 10¹7
2 item attempts remaining
Transcribed Image Text:The reaction 2 NO(g) + O2(g) → 2 NO2 (g) was studied, and the following data were obtained where Rate Δ[02] At [NO]。 [0₂]0 (molecules/cm³) (molecules/cm³) 1.00 × 10¹8 3.00 × 1018 2.50 × 1018 Submit Answer 1.00 × 10¹8 1.00 × 1018 2.50 × 1018 Initial Rate What would be the initial rate for an experiment where [NO]。 = 8.18 × 10¹8 molecules/cm³ and [0₂]0 = 3.57 × 10¹8 molecules/cm³? Rate = molecules/cm cm³ • S Try Another Version (molecules/cm³.s) 2.00 × 10¹6 1.80 × 10¹7 3.13 x 10¹7 2 item attempts remaining
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