The endpoint in tiration is the point where an indicatior's halfway thru its color change. Equivalence point is where moles/stoichiometry of the system is satisfied (moles of reactants are equal to each other).
Titrating a strong acid with a strong base results in a salt that is neutral. Phenolphthalein changes color in the range <span>8.3 – 10. It is very easy to spot the change as it is colorless in acidic (< 8.3) and pink in basic (> 10).
pH will rapidly change near titration equivalence point. </span><span>Only one drop of the titrant
causes this large change, the color change of phenolphthalein does not
occur on the equivalence point, but IT IS within about 1 drop. <span>It would be considered an "acceptable uncertainty" in using titration
to determine concentration by volumetric measurement.</span></span>
Answer:
.125 M
Explanation:
.15 M/L * .125 L = .01875 moles
now dilute to 150 cc (by adding 25 cc)
.01875M / (150/1000) = .125M
Titration experiments require the use of a burette. It is a long graduated glass tube held in place by a clamp stand. It has a tap fixture on the end that regulates the delivery of small volumes of liquid into a beaker in the titration process. Bunsen burners are used to heat substances and crucibles are used to hold items to be heated to high temperatures.
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
Answer:
B. velocity
Explanation:
Velocity is defined as speed in a given direction.
