The rate constant : k = 9.2 x 10⁻³ s⁻¹
The half life : t1/2 = 75.3 s
<h3>Further explanation</h3>
Given
Reaction 45% complete in 65 s
Required
The rate constant and the half life
Solution
For first order ln[A]=−kt+ln[A]o
45% complete, 55% remains
A = 0.55
Ao = 1
Input the value :
ln A = -kt + ln Ao
ln 0.55 = -k.65 + ln 1
-0.598=-k.65
k = 9.2 x 10⁻³ s⁻¹
The half life :
t1/2 = (ln 2) / k
t1/2 = 0.693 : 9.2 x 10⁻³
t1/2 = 75.3 s
In a place with no gravity
Explanation:
Weight is a force exerted by a body in the presence of gravity.
Weight = mass x gravity
Without gravity, there is no force of weight. The more the gravity, the more the weight of a body.
A body will experience weightlessness in a place without no gravity.
Therefore, one will experience the least weight in a place without the pull of gravity.
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weight and mass brainly.com/question/5956881
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Answer:
A: Synonyms
Explanation:
Synonyms are two words that have same meaning but are different words (tell me if this works!)
Answer:
<h2>8.5 g/cm³</h2>
Explanation:
The density of a substance can be found by using the formula

From the question we have

We have the final answer as
<h3>8.5 g/cm³</h3>
Hope this helps you
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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