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Evgen [1.6K]
3 years ago
6

CH3OH + O2 CO2 + H2O is this a combustion reaction?​

Chemistry
1 answer:
Brrunno [24]3 years ago
3 0

Answer:

yes

Explanation:

the results are CO2 and H2O, which usually appear after some kind of combustion.(combustion=burning)

hope this helps!

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How many grams would 3.36 × 1023 molecules of copper (II) sulfate (CuSO4) weigh?
Semenov [28]
Molar mass ( CuSO₄) = 159.609 g/mol

159.609 g ----------------- 6.02 x 10²³ molecules
         ? g ------------------ 3.36 x 10²³ molecules

mass = ( 3.36 x10²³) x 159.609 / 6.02 x 10²³

mass = 5.36 x 10²⁴ / 6.02 x 10²³

mass = 8.90 g 

hope this helps!
3 0
3 years ago
Read 2 more answers
What is the value of the equilibrium constant at 25 oC for the reaction between the pair: I2(s) and Br-(aq) Give your answer usi
insens350 [35]

Explanation:

Formula to calculate standard electrode potential is as follows.

          E^{o}_{cell} = E^{0}_{cathode} - E^{0}_{anode}

                             = 0.535 - 1.065

                             = - 0.53 V

Also, it is known that relation between E^{o}_{cell} and K is as follows.

            E^{o}_{cell} = \frac{RT}{nF} \times ln K

                 ln K = \frac{nFE^{0}_{cell}}{RT}      

Substituting the given values into the above formula as follows.

                 ln K = \frac{nFE^{0}_{cell}}{RT}    

                        =  \frac{2 \times 96485 C mol^{-1} \times -0.53 V}{8.314 l atm/mol K \times 298 K} \times \frac{1 J}{1 V C}  

                ln K = -41.28

                    K = e^{-41.28}    

                        = 1 \times 10^{-18}

Thus, we can conclude that the value of the equilibrium constant for the given reaction is 1 \times 10^{-18}.      

5 0
3 years ago
As a bond is broken in a compound the potential energy of the atoms generally does what
hichkok12 [17]
When a bond is broken that should be a type of reaction. When bonds are broken sometimes heat is released.
6 0
3 years ago
Which of the following would be one of the interacting parts in a weather system?
zhenek [66]

CLOUDS would be one of the interacting parts in a weather system.

6 0
3 years ago
The value of ka for nitrous acid (hno2) at 25 ∘c is 4.5×10−4. what is the value of δg at 25 ∘c when [h+] = 5.9×10−2m , [no2-] =
LuckyWell [14K]
To get the value of ΔG we need to get first the value of ΔG°:

when ΔG° = - R*T*㏑K

when R is constant in KJ = 0.00831 KJ

T is the temperature in Kelvin = 25+273 = 298 K

and K is the equilibrium constant = 4.5 x 10^-4

so by substitution:

∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4

        = -19 KJ

then, we can now get the value of ΔG when:

ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]

when ΔG° = -19 KJ

and R is constant in KJ = 0.00831 

and T is the temperature in Kelvin = 298 K

and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m  

so, by substitution:

ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )

       = -40

     
8 0
3 years ago
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