Answer:
m = 0.659 ounce
Explanation:
It is given that,
The thickness of a Teflon coating is, d = 1 mm
Area of the coating, A = 36 inch²
The density of Teflon, d = 0.805 g/mL
We need to find ounces of Teflon are needed.
Firstly, find the volume of the Teflon needed,
1 inch² = 6.4516 cm²
36 inch² = 232.258 cm²
Density,
V is volume of the Teflon needed, V = Ad
So,
Also, 1 gram = 0.035274 ounce
18.69 gram = 0.659 ounce
So, 0.659 ounces of Teflon are needed.
I believe the answer is: in order not to write very big or very small number values
It's adenosine triphosphate !
it has Penrose sugar and phosphate as backbone !
and nitrogenous base ... adenine.... in the middle !
Answer: ΔH for the reaction is -277.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%28products%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%28reactant%29%5D)
![\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCCl_4%7D%5Ctimes%20%5CDelta%20H_%7BCCl_4%7D%29%2B%28n_%7BHCl%7D%5Ctimes%20B.E_%7BHCl%7D%29%20%5D-%5B%28n_%7BCH_4%7D%5Ctimes%20%5CDelta%20H_%7BCH_4%7D%29%2Bn_%7BCl_2%7D%5Ctimes%20%5CDelta%20H_%7BCl_2%7D%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20-139%29%2B%281%5Ctimes%20-92.31%29%20%5D-%5B%281%5Ctimes%20-74.87%29%2B%281%5Ctimes%20121.0%5D)
Therefore, the enthalpy change for this reaction is, -277.4 kJ
I personally thinks it’s A
