Answer:
336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.
Explanation:
In this case, the balanced reaction is:
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:
- C₃H₈: 1 mole
- O₂: 5 moles
- CO₂: 3 moles
- H₂O: 4 moles
Being the molar mass of each compound:
- C₃H₈: 44 g/mole
- O₂: 16 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
Then, by stoichiometry, the following quantities of mass participate in the reaction:
- C₃H₈: 1 mole* 44 g/mole= 44 grams
- O₂: 5 moles* 16 g/mole= 80 grams
- CO₂: 3 moles* 44 g/mole= 132 grams
- H₂O: 4 moles* 18 g/mole= 72 grams
So you can apply the following rules of three:
- If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?
mass of CO₂= 336.6 grams
- If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?
mass of H₂O= 183.6 grams
<u><em>336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.</em></u>
Answer:
sorry man i cant help you i tried my best im so so so sorry
Explanation:
Answer:
C2= 0.16M
Explanation:
C1= 2M, V1= 20ml, C2= ?, V2= 250ml
Applying dilution formula
C1V1= C2V2
2×20 =C2×250
C2= 0.16M
Answer:
Atomic models are important because, they help us visualize the interior of atoms and molecules, and thereby predicting properties of matter.
Explanation:
We study the various atomic models in our course of study because, it is important for us to know, how did people come to the present concept of an atom. How did physics evolve from classical to quantum physics.
All these are important for us to know and thus, knowledge about various atomic models, their discoveries and drawbacks and finally improvements based on scientific evidence present at that time is important for us to understand the underlying theory very well.
Answer : The rate constant at 525 K is, 
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= ?
= activation energy for the reaction = 
R = gas constant = 8.314 J/mole.K
= initial temperature = 701 K
= final temperature = 525 K
Now put all the given values in this formula, we get:
![\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7B2.57M%5E%7B-1%7Ds%5E%7B-1%7D%7D%29%3D%5Cfrac%7B1.5%5Ctimes%2010%5E5J%2Fmol%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B701K%7D-%5Cfrac%7B1%7D%7B525K%7D%5D)
Therefore, the rate constant at 525 K is,
