According to this formula:
K= A*(e^(-Ea/RT) when we have K =1.35X10^2 & T= 25+273= 298K &R=0.0821
Ea= 85.6 KJ/mol So by subsitution we can get A:
1.35x10^2 = A*(e^(-85.6/0.0821*298))
1.35x10^2 = A * 0.03
A= 4333
by substitution with the new value of T(75+273) = 348K & A to get the new K
∴K= 4333*(e^(-85.6/0.0821*348)
= 2.16 x10^2
Answer:
<em>D. One negative charge</em>
Explanation:
During the formation of a bond, if an atom gains an electron, after that it will be left with a negative charge compared to the atom before the bond is formed. This is because in these types of bonds, which are <em>ionic bonds</em>, there is a <em>transfer of electrons between atoms</em>, there will be one or more atoms that yield electrons that will be captured by another and other atoms that gain them, and the difference of charges produced by this transfer of electrons, will cause the union to occur due to the attraction between electrostatic forces.
If you have a neutral atom before joining, and it gains an electron to form a bond,<em> it will have one electron more than its initial state</em> (in the initial state, the number of protons and electrons is the same, because the atoms they are electrically neutral), so having an extra electron will make it have a negative charge, since there will be a difference between the number of protons and electrons that the atom possesses. <em>This is why the correct answer is D.
</em>
In the case of <em>response A and B</em>, <em>the atom could only remain positively charged if it loses electrons</em>, but as in this case it wins, <em>they are not correct</em>.
<em>The answer C is also not correct</em> because only one electron wins, so that it is left with two negative charges, <em>it should gain two electrons during the bond formation.</em>
Answer:
the surface tension of H20 is 72 dynes/cm at 25°C
This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.
The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.
However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.
As an example, imagine two of the segmented lines have the following equations after applying the linear fit method:

First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:

Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:

This means the freezing point takes place after 7.72 second of cooling and is about 1.84 °C. Now you can replicate it for any not well-defined cooling curve.
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