**Question:**If you expect a chemical reaction to produce 4.38 L of oxygen gas at 19°C and 101 kPa, what will be the volume at 25°C and 98 kPa?**Explanation:**This problem involves using the Ideal Gas Law and the combined gas law formula to determine the change in volume of a gas under different conditions of temperature and pressure. **Solution Approach:**The combined gas law is:\[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \]Where:- \( P_1 \), \( V_1 \), \( T_1 \) are the initial pressure, volume, and temperature.- \( P_2 \), \( V_2 \), \( T_2 \) are the final pressure, volume, and temperature.- Temperatures need to be in Kelvin, so convert: \( T(K) = T(°C) + 273.15 \).1. Convert the temperatures from Celsius to Kelvin. - Initial temperature (\( T_1 \)): \( 19°C + 273.15 = 292.15 \, K \) - Final temperature (\( T_2 \)): \( 25°C + 273.15 = 298.15 \, K \)2. Substitute the given values into the combined gas law and solve for the unknown volume \( V_2 \):\[ \frac{101 \, \text{kPa} \times 4.38 \, \text{L}}{292.15 \, \text{K}} = \frac{98 \, \text{kPa} \times V_2}{298.15 \, \text{K}} \]Solve for \( V_2 \):\[ V_2 = \frac{101 \times 4.38 \times 298.15}{98 \times 292.15} \]Calculate \( V_2 \) to find the solution.

Given data set: V1= 4.38 L T1= 19 °C = 292 K P1 = 101 kPa V2= ? T2 = 25 °C = 298 K P2 = 98 kPaUsing equation: (P1*V1)/T1 = (P2*V2)/T2. Substituting known, (101 kPa*4.38 L)/292 K = (98 kPa*V2)/298 K V2= 4.60 L

Answered: emical reaction to produce 4.38 L of…

Science

Chemistry

emical reaction to produce 4.38 L of oxygen gas at th

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...

See similar textbooks

Similar questions

For a given amount of gas, the temperature varies O Inversely with volume and directly with pressure. O Inversely with volume and inversely with pressure. O Directly with volume and inversely with pressure. O Directly with volume and directly with pressure.
A gas initially at 2.5 L, 1.2 atm and 68 C undergoes a change so that its final volume and pressure are 1.6 L and 1.7 atm. Whats the final temperature??
Question 30 of 39 Submit Calculate the average kinetic energy for a mole of O2 at -23.0°C. Assume ideal gas behavior. J/mol 1 3 4 C 7 +/- x 10 0 Tap here or pull up for additional resources LO 00
A 144.6 mL sample of O, gas at 509.6 torr is compressed at constant temperature to 87.5 mL. Predict the pressure of the O, gas after this process. Put answer in box in regular non scientific notation with 1 place past decimal point
are their respective bulbs. When the valve is closed, the following 47. Consider two bulbs separated by a valve. Both bulbs are maintained at the same temperature. Assume that when the valve between the two bulbs is closed, the gases are sealed in O atm is pro- a 30.0% (by g/mL)? ive explosive aseous prod- nium nitrate valve between the two bulbs is closed, the gases are sealed data apply: Bulb A Bulb B CO Gas Ne the decom- 2.50 L 2.00 L V in a 50.0-L 1.09 atm 0.773 atm he resulting ? (Assume it Assuming no temperature change, determine the final inside the system after the valve connecting the two bulbs is opened. Ignore the volume of the tube connecting the two bulbs. 48. Follow the instructions of Problem 47 for the following set of bulbs. pressure but highly It can pass e presence rotoluene, Bulb X Bulb Y •mbustion producing Gas 0 CO2 CH4 tle when 5°C and 6.00 Lby V. 3.00 L P basgl 1.25 atm 2.66 atm
Use the following information to determine how many moles were produced? Round to two decimal places. Remember: for ideal gas law, you may need to convert multiple units. Pressure: = 742.1 mm HgVolume = 9.65 LTemperature = 45.5 °C
Only typed explanation otherwise leave it
otal pressure atm
Please don't provide handwritten solution..... A gas has a volume of 0.138 L at STP undergoes heating to and new pressure of 110.2 kPa. What is the new volume in liters ?