<u>Answer:</u> The value of 
for the net reaction is 13.94
<u>Explanation:</u>
The given chemical equations follows:
<u>Equation 1:</u> ![A+2B\xrightarrow[]{K_1} 2C](https://tex.z-dn.net/?f=A%2B2B%5Cxrightarrow%5B%5D%7BK_1%7D%202C)
<u>Equation 2:</u> ![2C\xrightarrow[]{K_2} D](https://tex.z-dn.net/?f=2C%5Cxrightarrow%5B%5D%7BK_2%7D%20D)
The net equation follows:
![D\xrightarrow[]{K_c} A+2B](https://tex.z-dn.net/?f=D%5Cxrightarrow%5B%5D%7BK_c%7D%20A%2B2B)
As, the net reaction is the result of the addition of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the inverse of second equilibrium constant.
The value of equilibrium constant for net reaction is:
We are given:
Putting values in above equation, we get:
Hence, the value of for the net reaction is 13.94
☁️ Answer ☁️
annyeonghaseyo!
Your answer is:
it can bond with oxygen to form rust or iron oxide. and it can bond with carbon to form steel.
Here's another one: Iron bonds with lots of things: Oxygen (Ferric Oxide = rust),Chlorine (Ferric Chloride]), Fluorine, ... , even other Iron (in aMetallic Bond or crystal).
Hope it helps.
Have a nice day hyung/noona!~ ￣▽￣❤️
Answer:
Explanation:
a) In an exothermic reaction, the energy transferred to the surroundings from forming new bonds is ___more____ than the energy needed to break existing bonds.
b) In an endothermic reaction, the energy transferred to the surroundings from forming new bonds is ___less____ than the energy needed to break existing bonds.
c) The energy change of an exothermic reaction has a _____negative_______ sign.
d) The energy change of an endothermic reaction has a ____positive________ sign.
The energy changes occur during the bonds formation and bonds breaking.
There are two types of reaction endothermic and exothermic reaction.
Endothermic reactions:
The type of reactions in which energy is absorbed are called endothermic reactions.
In this type of reaction energy needed to break the bond are higher than the energy released during bond formation.
For example:
C + H₂O → CO + H₂
ΔH = +131 kj/mol
it can be written as,
C + H₂O + 131 kj/mol → CO + H₂
Exothermic reaction:
The type of reactions in which energy is released are called exothermic reactions.
In this type of reaction energy needed to break the bonds are less than the energy released during the bond formation.
For example:
Chemical equation:
C + O₂ → CO₂
ΔH = -393 Kj/mol
it can be written as,
C + O₂ → CO₂ + 393 Kj/mol
Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
when we have Pv(solu) = 22.8 and Pv(water) = 23.8 so by substitution:
22.8 = n * 23.8
n= 0.958
- we need to get the moles of glucose:
moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958 = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
0.968 moles of glucose = 1.2
moles of glucose = 1.253 mol
∴ the mass of glucose = no.of glucose moles x molar mass
= 1.253 x 180 = 225.5 g
Q: B
here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
n = 0.88
so, mole fraction of solution = 1 - 0.88 = 0.12
and we can get after that the moles of water = (mass weight / molar mass)
- no.moles of water = 85 g / 18 g/mol = 4.7 moles
- total moles in solution = moles of water / moles fraction of water
= 4.7 / 0.88 = 5.34 moles
∴ moles of the solution = total moles in solu - moles of water
= 5.34 - 4.7 = 0.64 moles solute
∴ the molar mass of the solute = mass weight of solute / no.of moles of solute
= 53.8 / 0.64 = 84 g/mole
Q: C
moles of urea (NH2)2 CO = mass weight / molar mass
= 4.49 g / 60 g /mol
= 0.07 mol
moles of methanol = mass weight / molar mass
= 39.9 g / 32 g/mol = 1.25 mol
moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
moles fraction of methanol = 1.25 / ( 1.25+0.07) = 0.95
by substitution in Pv formula we will be able to get the vapour pressure of the solu :
Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg
∴Pv(solu) = 84.55 mmHg
