Answer:
Will group 2 elements lose electrons to bond with nonmetals in group 17 in a
1:2 ratio?
Explanation:
Metals are electropositive in nature. This means that they loose electrons. Thus, metals form ionic bonds by loosing electrons to non metals.
Elements of group 2 have a valency of 2 while those of group 17 has a valency of 1 so the ratio in which group 2 elements bond with elements of group 17 is 1:2. Hence the answer.
Answer:
it will sink to the bottom
Explanation:
if it has a higher density than the water then it will sink.
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
#SPJ4.
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
The event that is known to have 'created' the universe was the Big Bang THEORY (because it can't be proven and most people just don't want to admit that there is a God)
Answer : The equilibrium concentration of CO in the reaction is, 
Explanation :
The given chemical reaction is:
The expression for equilibrium constant is:
![K_c=\frac{[COCl_2]}{[CO][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCOCl_2%5D%7D%7B%5BCO%5D%5BCl_2%5D%7D)
As we are given:
Concentration of 
at equilibrium = Concentration of 
So,
![K_c=\frac{[Cl_2]}{[CO][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCl_2%5D%7D%7B%5BCO%5D%5BCl_2%5D%7D)
![K_c=\frac{1}{[CO]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B1%7D%7B%5BCO%5D%7D)
![1.2\times 10^3=\frac{1}{[CO]}](https://tex.z-dn.net/?f=1.2%5Ctimes%2010%5E3%3D%5Cfrac%7B1%7D%7B%5BCO%5D%7D)
![[CO]=8.3\times 10^{-4}M](https://tex.z-dn.net/?f=%5BCO%5D%3D8.3%5Ctimes%2010%5E%7B-4%7DM)
Therefore, the equilibrium concentration of CO in the reaction is,
