<u>Answer:</u> The correct answer is 1.18 g.
<u>Explanation:</u>
We are given a chemical equation:
We know that at STP conditions:
22.4L of volume is occupied by 1 mole of a gas.
So, 2.21L of carbon dioxide is occupied by = 
of carbon dioxide gas.
By Stoichiometry of the above reaction:
1 mole of carbon dioxide gas is produced by 1 mole of carbon
So, 0.0986 moles of carbon dioxide is produced by = 
of carbon.
Now, to calculate the mass of carbon, we use the equation:
Moles of carbon = 0.0986 mol
Molar mass of carbon = 12 g/mol
Putting values in above equation, we get:
Hence, the correct answer is 1.18 g.
First, the sun shines liquid (ocean) Next, the water evaporates
Let us assume that this molecule is 100 percent ionic. In that case, the charges are distinguished by a bond length.
h = Q × r
= (160 × 10⁻¹⁹ c) (127 × 10⁻¹² m) (10 / 3.336 × 10⁻³⁰ cm)
= 6.09 D.
The actual dipole moment is = 1.08 D
Therefore, the percent ionic character is,
= 1.08 D / 6.09 D × 100
= 17.7 %.
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
Make sure that you understand what they are asking you from this question, as it can be confusing, but the solution is quite simple. They are stating that they want you to calculate the final concentration of 6.0M HCl once a dilution has been made from 2.0 mL to 500.0 mL. They have given us three values, the initial concentration, initial volume and the final volume. So, we are able to employ the following equation:
C1V1 = C2V2
(6.0M)(2.0mL) = C2(500.0mL)
Therefore, the final concentration, C2 = 0.024M.
