Answer:
The answer to your question is: 69.6 %
Explanation:
Freon -112 (C₂Cl₄F₂)
MW = (12 x 2) + (35.5 x 4) + (19 x 2)
= 24 + 142 + 38
= 204 g
204 g of C₂Cl₄F₂ ----------------- 100%
142 g ----------------- x
x = (142 x 100 ) / 204
x = 69.6 %
Answer:
The pH of the buffer is 7.0 and this pH is not useful to pH 7.0
Explanation:
The pH of a buffer is obtained by using H-H equation:
pH = pKa + log [A⁻] / [HA]
<em>Where pH is the pH of the buffer</em>
<em>The pKa of acetic acid is 4.74.</em>
<em>[A⁻] could be taken as moles of sodium acetate (14.59g * (1mol / 82g) = 0.1779 moles</em>
<em>[HA] are the moles of acetic acid (0.060g * (1mol / 60g) = 0.001moles</em>
<em />
Replacing:
pH = 4.74 + log [0.1779mol] / [0.001mol]
<em>pH = 6.99 ≈ 7.0</em>
<em />
The pH of the buffer is 7.0
But the buffer is not useful to pH = 7.0 because a buffer works between pKa±1 (For acetic acid: 3.74 - 5.74). As pH 7.0 is out of this interval,
this pH is not useful to pH 7.0
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Answer:
By visiting other households with cats.
Explanation:
This will give Brian a variety of other houses and determine if it is truly cats or just alleries from other items. This is the most direct way to get Brian the answer he is looking for.
Answer:
Explanation:
Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.
In the living E. coli cells,
[ATP] = 7.9 mM;
[ADP] = 1.04 mM,
[glucose] = 2 mM,
[glucose 6-phosphate] = 1 mM.
Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
The reaction is given as
Glucose + ATP → glucose 6-phosphate + ADP
Now reaction quotient for given equation above is
![q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B%5B%5Ctext%20%7Bglucose%206-phosphate%7D%5D%5BADP%5D%7D%7B%5BGlucose%5D%5BATP%5D%7D)
so,
⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq
