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
D. A sour liquid that forms gas bubblee when mixed with copper
the value of delta H° for aniline = -3201. 5 Kj/mol
<u><em> calculation</em></u>
Step 1: find heat
Q ( heat) = C ( specific heat capacity) x ΔT ( change in temperature)
C= 4.25 kj/c°
ΔT = 69.8-29.5 = 40.3 c°
Q= 4.25 kj/c x 40.3 c = 171.28 kj
Step 2: find the moles of aniline
moles = mass/ molar mass
= 4.98 g/ 93.13 g/mol =0.0535 moles
Step 3 : find delta H
171.28 kj / 0.0535=3201.5 kj/mol
since the reaction is exothermic delta H = -3201.5 Kj/mol
Answer:
When electromagnetic radiations enter earth surface , it passes through several layers of our atmosphere in meantime some raditions get reflected back and some reaches to us .
-dB/dt = kAB = k(2B)(B) = 2kB^2
-dB/B^2 = 2kdt
Integrating: 1/B - 1/(B_0) = 2kt
At t = 10, if 15 g of C have formed, this must have consumed 10 g A and 5 g B. The remaining mass of B is 45 g.
1/45 - 1/50 = 2(k)(10)
k = 1.11 x 10^-4
Then substituting this value of k with t = 40:
1/B - 1/50 = 2(1.11 x 10^-4)(40)
1/B - 1/50 = 0.008889
1/B = 0.028889
B = 34.62 g remaining
Therefore, 50 - 34.62 = 15.38 g of B have been consumed.
Doubling, 30.76 g of A have been consumed.
This means that 15.38 + 30.76 = 46.15 g of C have been formed.
