You're a chemistry angel praise you and your family thank you so much ❤❤❤ my school is in the weeds for science courses and youre just an unsung savior
To solve the problem of finding the enthalpy change when 25.0 grams of ammonia reacts with excess oxygen, follow these steps: 1. Begin with the balanced chemical equation and the given enthalpy change: 2 NH3(g) + 2 O2(g) → N2O(g) + 3 H2O(l) with ΔH = -683 kJ. 2. Convert the mass of NH3 to moles. Using the molar mass of NH3 (17.03 g/mol), divide the mass by the molar mass: 25.0 g ÷ 17.03 g/mol = 1.468 moles of NH3. 3. Determine the enthalpy change for the given moles of NH3. The given enthalpy change (-683 kJ) corresponds to the reaction of 2 moles of NH3. Set up a proportion to find the enthalpy change for 1.468 moles: ΔH = (-683 kJ / 2 moles) × 1.468 moles. 4. Calculate the enthalpy change: ΔH = -683 kJ ÷ 2 × 1.468 = -500.94 kJ. 5. Thus, the enthalpy change when 25.0 grams of NH3 reacts with excess O2 is approximately -500.94 kJ. Hope that helps! Thanks for watching!
@48:24, 827/2=413.5 not 436! Sorry!
you don't know how much you've helped me with my chem class, thank you so much
That makes me so happy, I’m so glad I can help!! Rooting for you queen!!!
You're a chemistry angel praise you and your family thank you so much ❤❤❤ my school is in the weeds for science courses and youre just an unsung savior
Omg thank you so much for your kind words!!!! This brightened my day, much love and good luck with your class 💖
May Allah reward you Professor for your effort! Waiting for new series.
Many many thanks🥰❤️
48:43 I found it -468 kj. However ı couldn't understand what is w/xs meaning?
With excess = w/xs. Sorry for not making that clear
@@professoreman2289thank you for feedback. İ Found the correct answer when I change the calculater from phone to scientific calculater.
Eman! It's Emily and you're awesome!
EMILY!! Thank you for the support always ❤️
huge respect to you
Thank you so much ❤️
May Allah bless you for this
Thank you so so much for watching, please keep me in your duas ❤️
how do you solve example #5? love all your videos
To solve the problem of finding the enthalpy change when 25.0 grams of ammonia reacts with excess oxygen, follow these steps:
1. Begin with the balanced chemical equation and the given enthalpy change: 2 NH3(g) + 2 O2(g) → N2O(g) + 3 H2O(l) with ΔH = -683 kJ.
2. Convert the mass of NH3 to moles. Using the molar mass of NH3 (17.03 g/mol), divide the mass by the molar mass: 25.0 g ÷ 17.03 g/mol = 1.468 moles of NH3.
3. Determine the enthalpy change for the given moles of NH3. The given enthalpy change (-683 kJ) corresponds to the reaction of 2 moles of NH3. Set up a proportion to find the enthalpy change for 1.468 moles: ΔH = (-683 kJ / 2 moles) × 1.468 moles.
4. Calculate the enthalpy change: ΔH = -683 kJ ÷ 2 × 1.468 = -500.94 kJ.
5. Thus, the enthalpy change when 25.0 grams of NH3 reacts with excess O2 is approximately -500.94 kJ.
Hope that helps! Thanks for watching!
Yes, thanks I got the moles 1.468 of NH3, but I got stuck on what to do next.
What books is this going of off ?
She said it's based on multiple authors but primarily Jimmy Roger's General Chemistry Textbook.
yes and zumdahls!
do you have a practice problem set video for this chapter
You can find some more problems in the MCAT Chem playlist too
@@professoreman2289 thank you so much! I love your videos, they are soooo helpful!!
We would appreciate if you Professor uploap more because final exams are coming(
I’ll try my best
Good luck on your finals!
@@emilyg2451 Thank you!
Hi! BTW I already have a completed gen chem 1 and 2 playlist. if videos are not uploaded that you need to study, check those out in the time being
@@professoreman2289 Thank you Professor
İs tehere any general physics chanel's like this
I’m hoping to make one soon!