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The heat released by cpmolete the combustion of organic products with oxygen is called heat of combustion.
Here 1 mol of CH4 realesed 802.3 KJ
To emit 264 kJ you multiply you need (1 mol of CH4/802.3 kJ)* 264 kJ = 0.329 mol of CH4
The molar mass, MM, of CH4 is 12 g/mol + 4*1g/mol = 16 g/mol
The to obtain the mass multiply the number of moles times the molar mass:
mass = n * MM = 0.329mol * 16g/mol = 5.26 grams
Answer: 5.26 grams
Answer:
Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)
Explanation:
Let's consider the molecular single displacement equation between Zn and Cu(NO₃)₂
Zn(s) + Cu(NO₃)₂(aq) ⇒ Zn(NO₃)₂(aq) + Cu(s)
The complete ionic equation includes all the ions and insoluble species.
Zn(s) + Cu²⁺(aq) + 2 NO₃⁻(aq) ⇒ Zn²⁺(aq) + 2 NO₃⁻(aq) + Cu(s)
The net ionic equation includes only the ions that participate in the reaction and insoluble species.
Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)
Answer: The hydroboration of an alkene occurs in TWO CONCERTED STEP which places the boron of the borane on the LESS SUBSTITUTED carbon of the double bond. The oxidizing agent then acts as a nucleophile, attacking the electrophilic BORON and resulting in the placement of a hydroxyl group on the attached carbon. Thus, the major product of the hydroboration oxidation reaction DOES NOT follow Markovnikov's rule.
Explanation:
Hydroboration is defined as the process which allows boron to attain the octet structure. This involves a two steps pathway which leads to the production of alcohol.
--> The first step: this involves the initiation of the addittion of borane to the alkene and this proceeds as a concerted reaction because bond breaking and bond formation occurs at the same time.
--> The second step: this involves the addition of boron which DOES NOT follow Markovnikov's rule( that is, Anti Markovnikov addition of Boron). This is so because the boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon.
Note: The Markovnikov rule in organic chemistry states that in alkene addition reactions, the electron-rich component of the reagent adds to the carbon atom with fewer hydrogen atoms bonded to it, while the electron-deficient component adds to the carbon atom with more hydrogen atoms bonded to it.
The question is incomplete, complete question is :
The first two steps in the industrial synthesis of nitric acid produce nitrogen dioxide from ammonia:
Step 1 : 
Step 2 : 
The net reaction is:

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants
and
. If you need to include any physical constants, be sure you use their standard symbols
Answer:
Equation that gives the overall equilibrium constant K in terms of the equilibrium constants
:

Explanation:
Step 1 : 
Expression of an equilibrium constant can be written as:
![K_1=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BNO%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E5%7D)
Step 2 : 
Expression of an equilibrium constant can be written as:
![K_2=\frac{[NO_2]^2}{[NO]^2[O_2]}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BNO%5D%5E2%5BO_2%5D%7D)
The net reaction is:

Expression of an equilibrium constant can be written as:
![K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D)
Multiply and divide
;
![K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO]^4}{[NO]^4}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D%5Ctimes%20%5Cfrac%7B%5BNO%5D%5E4%7D%7B%5BNO%5D%5E4%7D)
![K=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO_2]^4}{[NO]^4}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D%5Ctimes%20%5Cfrac%7B%5BNO_2%5D%5E4%7D%7B%5BNO%5D%5E4%7D)
![K=K_1\times \frac{[NO_2]^4}{[O_2]^2[NO]^4}](https://tex.z-dn.net/?f=K%3DK_1%5Ctimes%20%5Cfrac%7B%5BNO_2%5D%5E4%7D%7B%5BO_2%5D%5E2%5BNO%5D%5E4%7D)
![K=K_1\times (\frac{[NO_2]^2}{[O_2]^1[NO]^2})^2](https://tex.z-dn.net/?f=K%3DK_1%5Ctimes%20%28%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BO_2%5D%5E1%5BNO%5D%5E2%7D%29%5E2)

So , the equation that gives the overall equilibrium constant K in terms of the equilibrium constants
:
