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Acidic and basic are two extremes that describe chemicals, just like hot and cold are two extremes that describe temperature. Mixing acids and bases can cancel out their extreme effects, much like mixing hot and cold water can even out the water temperature. A substance that is neither acidic nor basic is neutral.
The character of acidic, basic and neutral is defined by the concentration of hydrogen ions [H+](mol/L). A solution with a concentration of hydrogen ions higher than 10-7mol/L is acidic, and a solution with a lower concentration is alkaline (another way to say basic). Using the formula, pH=-log[H+], a pH of 7 is neutral, a pH less than 7 is acidic, and a pH greater than 7 is basic. As one can see from this formula, ten times a given concentration of hydrogen ions means one unit lower in terms of pH value (higher acidity), and vice versa.
The formula for ph is given by:pH=−log10[H+]
What is the concentration of H+ ions at a pH = 8?
In calculating for the concentration of hydrogen ion, the formula is given by:[H+]=(10)^(-pH)
Solution:
[H+]=(10)^(-8)[H+]=0.00000001 mol/L
What is the concentration of OH– ions at a pH = 8?pH+pOH=148+pOH=14pOH=6
[OH-]=(10)^(-pOH)[OH-]=(10)^(-6)[OH-]=0.000001
What is the ratio of H+ ions to OH– ions at a pH = 2?The ratio is 0.00000001:0.000001 which is equal to 0.01
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
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