Answer : The limiting reagent is 
Solution : Given,
Moles of methane = 2.8 moles
Moles of = 5 moles
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of 
So, 5 moles of react with 
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Hence, the limiting reagent is
Answer:
No energy is gained or lost when molecules collide. The molecules in a gas take up a negligible (able to be ignored) amount of space in relation to the container they occupy. The molecules are in constant, linear motion.
Explanation:
Hope this helps!
Answer:
Enthalpy change for the reaction is -67716 J/mol.
Explanation:
Number of moles of 
in 50.0 mL of 0.100 M of
= Number of moles of HCl in 50.0 mL of 0.100 M of HCl
= 
moles
= 0.00500 moles
According to balanced equation, 1 mol of reacts with 1 mol of HCl to form 1 mol of AgCl.
So, 0.00500 moles of react with 0.00500 moles of HCl to form 0.00500 moles of AgCl
Total volume of solution = (50.0+50.0) mL = 100.0 mL
So, mass of solution = (
) g = 100 g
Enthalpy change for the reaction = -(heat released during reaction)/(number of moles of AgCl formed)
= 
= ![\frac{-100g\times 4.18\frac{J}{g.^{0}\textrm{C}}\times [24.21-23.40]^{0}\textrm{C}}{0.00500mol}](https://tex.z-dn.net/?f=%5Cfrac%7B-100g%5Ctimes%204.18%5Cfrac%7BJ%7D%7Bg.%5E%7B0%7D%5Ctextrm%7BC%7D%7D%5Ctimes%20%5B24.21-23.40%5D%5E%7B0%7D%5Ctextrm%7BC%7D%7D%7B0.00500mol%7D)
= -67716 J/mol
[m = mass, c = specific heat capacity, 
= change in temperature and negative sign is included as it is an exothermic reaction]
Hello!
The volume of the gas when its pressure is increased to 880 mm Hg is
88,64 mLThis question can be easily answered using the
Boyle's Law, which states that as pressure increases, the volume is lower. It can be expressed in a mathematical way as follows:
So, from this equation we can clear V₂ to find the volume at 880 mm Hg:
Have a nice day!
We are using the General gas equation P x V/K = P x V/K
1. P = 1atm V=500ml so PxV= 500 at 6.5km P = 0.5atm V = ? so P xV = 0.5 x V
(We don't have to worry about temperature!) 500 = 0.5 x V so V = 1000ml
2. NO CHANGE in pressure here so we have V/K V=2.75 K = 20 + 273=293 so V/K= 2.75/293
Next set V = 2.46 K = ? so V/K = 2.46/K then 2.75/293= 2.46/K so K=(293/2.75)x2.46
=262 K
Convert back to Celsius 262 - 273 = -11 C
It's raining so I have to rescue the laundry!
Laundry rescued!
3.Now we have to use all three variables. I am using 273K and 100kPa for STP.
P = 100 V = 700 K = 273 These are altered P - unknown, V = 200 K = 273+30=303
!00 x 700/273 = 256.4 this is equal to P x 200/303 = P x 0.66
so P = 256.4/0.66 = 388.48kPa
