Answer:
<em>The type of vegetation a surface does affect the </em><em>water coming from above to sink in or runoff. </em>
Explanation:
This is how the vegetation affects the runoff:-
The leaves and stems present in the vegetation do not let the water fall directly on the soil and makes the process rather slow which makes the water to get to the ground slowly and sink in properly inside the soil rather than running off.
If the vegetation present is dense with there was being hairy then also the water would not run out and will get absorbed by the roots letting the soil intact
The correct answer is: [C]:
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"<span>pressure and the number of gas molecules are directly related."
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<u>Note</u>: The conclusion was: "</span> as the pressure in a system increases, the number of gas molecules increases" — over the course of many trials.
This means that the "pressure" and the "number of gas molecules" are directly related.
Furthermore, this conclusion is consistent with the "ideal gas law" equation:
" PV = nRT " ;
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in which:
"P = Pressure" ;
"n = number of gas molecules" ;
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All other factors held equal, when "n" (the "number of gas molecules")
increases in value (on the "right-hand side" of the equation), the value for "P" (the "pressure" — on the "left-hand side" of the equation), increases.
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<span>Answer:
Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either.
Start with your 2 half reactions:
I- --> IO3-
Cr2O72- --> 2 Cr3+
Balance O by adding H2O:
I- + 3 H2O --> IO3-
Cr2O72- --> 2 Cr3+ + 7H2O
Balance H by adding H+:
I- + 3 H2O --> IO3- + 6 H+
Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O
Balance charge by adding e-:
I- + 3 H2O --> IO3- + 6 H+ + 6 e-
Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O
Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give:
Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O</span>
<u>Answer:</u> The pH of resulting solution is 8.7
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
Molarity of TRIS acid solution = 0.1 M
Volume of solution = 50 mL
Putting values in above equation, we get:
Molarity of TRIS base solution = 0.2 M
Volume of solution = 60 mL
Putting values in above equation, we get:
Volume of solution = 50 + 60 = 110 mL = 0.11 L (Conversion factor: 1 L = 1000 mL)
- To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
![pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5B%5Ctext%7BTRIS%20base%7D%5D%7D%7B%5B%5Ctext%7BTRIS%20acid%7D%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of TRIS acid = 8.3
![[\text{TRIS acid}]=\frac{0.005}{0.11}](https://tex.z-dn.net/?f=%5B%5Ctext%7BTRIS%20acid%7D%5D%3D%5Cfrac%7B0.005%7D%7B0.11%7D)
![[\text{TRIS base}]=\frac{0.012}{0.11}](https://tex.z-dn.net/?f=%5B%5Ctext%7BTRIS%20base%7D%5D%3D%5Cfrac%7B0.012%7D%7B0.11%7D)
pH = ?
Putting values in above equation, we get:
Hence, the pH of resulting solution is 8.7
