Sample means for solutions 1 and 2 are 19.27 and 10.32 respectively
In semiconductor manufacturing,
The total for answer 1 is given by:
9.7+10.5+9.4+10.6+9.3+10.7+9.6+10.4+10.2+10.5 = 192.7
The sample size is 10 and provides us with
192.7/10 = 19.27
For solution 2, the sum is given by:
10.6+10.3+10.3+10.2+10.0+10.7+10.3+10.4+10.1+10.3 = 103.2
The sample size is 10, this gives us
103.2/10 = 10.32
The total for answer 2 is given by:
10.6+10.3+10.3+10.2+10.0+10.7+10.3+10.4+10.1+10.3 = 103.2
The sample size is 10 and provides us with
103.2/10 = 10.32
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In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the backs of wafers prior to metalization. The etch rate is an important characteristic in this process and is known to follow a normal distribution. Two different etching solutions have been compared, using two random samples of 10 wafers for each solution. Assume the variances are equal. The etch rates are as follows (in mils per minute): Solution 1 Solution 2 9.7 10.6 10.5 10.3 9.4 10.3 10.6 10.2 9.3 10.0 10.7 10.7 9.6 10.3 10.4 10.4 10.2 10.1 10.5 10.3 Calculate sample means of solution 1 and solution 2
Explanation:
When a strong acid, say
reacts which a weak base, say
, the reaction is shown below as:-

The salt further reacts with water as shown below:-

Formation of
lowers the pH value of the solution as more hydrogen ions leads to less pH.
This question is asking for the pH of a buffer solution between ammonia and nitric acid, with given volumes and concentrations. At the end, the result turns out to be 10.488.
<h3>Buffers</h3>
In chemistry, buffers are known as substances attempting to hold a relatively constant pH by mixing and acid and a base (weak and strong). In such a way, for the substances given, the first step will be to calculate the consumed moles as they are mixed:

Now, since ammonia is in a greater proportion, one can calculate how much of it is left after being consumed by the nitric acid:

And its new concentration:
![[NH_3]=\frac{0.001mol}{0.1L+0.09L} =0.00526M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D%5Cfrac%7B0.001mol%7D%7B0.1L%2B0.09L%7D%20%3D0.00526M)
Next, with ammonia's ionization:

We set up the equilibrium expression based on ammonia's Kb:
![Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D)
Which can be solved by introducing x and using ammonia's Kb:

Then, we solve for x which is also equal to the concentration of ammonium and hydroxide ions in the solution:

Ultimately, we calculate the pOH and then turn it into pH with:

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