Answer:
The new volume at -29°C and 432 torr is 6.51 L
Explanation:
Step 1: Data given
Initial volume of CO = 4.63 L
Initial temperature = 296.0 K
Initial pressure = 737 torr = 737/760 = 0.9697 atm
Step 2: Calculate the new volume
(P1*V1)/T1 = (P2*V2)/T2
V2 = (P1V1*T2) / (T1*P2)
⇒ with P1 = the initial pressure = 0.9697 atm
⇒ with V1 = the initial volume = 4.63 L
⇒ with T1 = The initial temperature = 296.0 K
⇒ with T2 = the new temperature = -29.0 + 273 = 244 K
⇒ with P2 = the new pressure = 432/760 = 0.5684 atm
V2 = (P1V1/T1) * (T2/P2)
V2 = 6.51 L
The new volume at -29°C and 432 torr is 6.51 L
Answer:
the awnser is 22
Explanation:
9+6 = 15 + 7 = 22 have a great day
Answer:
The reaction in this experiment is termed an iodine clock reaction, because it is the molecular iodine (I2) that undergoes the sudden concentration change. When the iodine concentration increases, it reacts with the starch in the solution to form a complex, turning it a deep blue-black color.
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
Learn more about semiconductor manufacturing here brainly.com/question/22779437
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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
