Answer:
CO + 2H2 = CH3OH
Explanation:
1. Label Each Compound With a Variable
aCO + bH2 = cCH3OH
2. Create a System of Equations, One Per Element
C: 1a + 0b = 1c
O: 1a + 0b = 1c
H: 0a + 2b = 4c
3. Solve For All Variables (using substitution, gauss elimination, or a calculator)
a = 1
b = 2
c = 1
4. Substitute Coefficients and Verify Result
CO + 2H2 = CH3OH
L R
C: 1 1 ✔️
O: 1 1 ✔️
H: 4 4 ✔️
Answer:
Molecular solids and covalent network solids are two types of solid compounds. The key difference between molecular solid and covalent network solid is that <em>molecular solid forms due to the action of Van der Waal forces </em>where as <em>covalent network solid forms due to the action of covalent chemical bonds.</em>
hope this helps
Answer: 116g/mole
Explanation:
She didn't get the answer because she didn't add the them well , due to the bracket present.
Answer : The equilibrium constant for this reaction is, 
Explanation :
The given main chemical reaction is:
; 
The intermediate reactions are:
(1)
; 
(2)
; 
We are reversing reaction 1 and multiplying reaction 2 by 2 and then adding both reaction, we get:
(1)
; 
(2)
; 
Thus, the equilibrium constant for this reaction will be:


Thus, the equilibrium constant for this reaction is, 