**Problem Statement:**Find $ v_{\text{rms}} $ for nitric oxide gas (NO) at 25°C.**Periodic Table Information:**- Oxygen (O): Atomic number: 8, Atomic mass: 16.00- Nitrogen (N): Atomic number: 7, Atomic mass: 14.01**Answer Choices:**A. $ 497.8 \, \text{m/s} $B. $ 600.1 \, \text{m/s} $C. $ 405.0 \, \text{m/s} $D. $ 527.3 \, \text{m/s} $E. $ 594.7 \, \text{m/s} $F. $ 403.7 \, \text{m/s} $

Answered: Find v,ms for nitric oxide gas (NO) at…

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

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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**Problem Statement:**

Find $ v_{\text{rms}} $ for nitric oxide gas (NO) at 25°C.

**Periodic Table Information:**

- Oxygen (O): Atomic number: 8, Atomic mass: 16.00
- Nitrogen (N): Atomic number: 7, Atomic mass: 14.01

**Answer Choices:**

A. $ 497.8 \, \text{m/s} $

B. $ 600.1 \, \text{m/s} $

C. $ 405.0 \, \text{m/s} $

D. $ 527.3 \, \text{m/s} $

E. $ 594.7 \, \text{m/s} $

F. $ 403.7 \, \text{m/s} $

Expert Solution

Step 1 To Calculate

$G i v e n, t e m p e r a t u r e = 25 ° C = (25 + 273.15) K = 298.15 K M o l a r m a s s o f n i t r i c o x i d e = 30.01 g / m o l = 0.03001 k g / m o l g a s c o n s \tan t = 8.314 \frac{J}{m o l . K} w e a r e a s k e d t o c a l c u l a t e t h e V_{r m s} f o r n i t r i c o x i d e a t 25 ° C$

Step 2 Formula used and Calculations

$T h e V_{r m s} o f n i t r i c o x i d e g a s i s g i v e n b y, V_{r m s} = \sqrt{\frac{3 R T}{M}}; w h e r e, R i s g a s c o n s \tan t, T i s t e m p e r a t u r e a n d M i s m o l a r m a s s o f g a s . N o w, V_{r m s} = \sqrt{\frac{3 R T}{M}} V_{r m s} = \sqrt{\frac{3 \times 8.314 \frac{J}{m o l . K} \times 298.15 K}{0.03001 \frac{k g}{m o l}}} V_{r m s} = \sqrt{\frac{7436.4573 \frac{k g . m^{2}}{s e c^{2}}}{0.03001 k g}} (1 J = 1 \frac{k g . m^{2}}{s e c^{2}}) V_{r m s} = \sqrt{247799.3102 \frac{m^{2}}{s e c^{2}}} V_{r m s} = 497.794 \frac{m}{s e c} V_{r m s} = 497.8 \frac{m}{s e c} T h u s, t h e V_{r m s} f o r n i t r i c o x i d e a t 25 ° C w o u l d b e, 497.8 \frac{m}{s e c}$

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