Copper choruses purpose is its used as a catalyst for organic and inorganic reactions , mordant for dyeing and printing textiles, pigment for glass and ceramics, wood preservative, disinfectant, insecticide, fungicide, and herbicide. (Not sure if you want the actual purpose of water or not)
I don't know what the problem is, but here are some rues to help you out:
- All non-zero figures are significant
- When a zero falls between non-zero digits, that zero is significant.
- When a zero falls after a decimal point, that zero is significant.
- When multiplying and dividing significant figures, the answer is limited to the number of sig figs equal to the least number of sig figs in the problem.
- When adding and subtracting, the answer is limited to the number of decimal places in the number with the least number of decimal places.
Ethane is an alkane. Methane is also an alkane and is considered to be the simplest alkane. The difference is ethane has only 2 carbon. That carbon has 6 hydrogen attached to it. So what we do is we multiply the moles of ethane by the number of hydrogen (by dimension analysis) resulting to 82.68 moles H.
The molecular formula :
C₆H₁₄O₃PF
<h3>Further explanation</h3>
Given
39.10% carbon, 7.67% hydrogen, 26.11% oxygen, 16.82% phosphorus, and 10.30% fluorine.
Required
The molecular formula
Solution
mol ratio :
C = 39.1 : 12 = 3.258
H = 7.67 : 1 = 7.67
O = 26.11 : 16 = 1.632
P = 16.82 : 31 = 0.543
F = 10.3 : 19 = 0.542
Divide by 0.542
C = 6
H : 14
O = 3
P = 1
F = 1
The empirical formula :
C₆H₁₄O₃PF
(The empirical formula)n = the molecular formula
(C₆H₁₄O₃PF)=184.1
(6.12+14.1+3.16+31+19)n=184.1
(184)n=184.1
n = 1
The range is negative numbers.
The interval for the range is .
***You might want to look at your functions again because I don't see a choice that matches.
Step-by-step explanation:
Given functions:
We are asked to find the range of .
I'm also going to look at the domain just to see if this possibly might change my range .
is the inner function. So we will consider the domain of that function first.
You only have to worry about division by zero for the function .
Since we are dividing by , we don't want to be zero.
So far the domain is all real numbers except .
Now let's move out.
exists for all numbers, . So we didn't want to include from before.
Now let's put it together:
So the domain is still all real numbers except at since we cannot divide by 0 and is 0 when .
with .
is positive for all numbers except .
So is negative for all numbers since negative divided by positive is negative.
So the range is only negative numbers.
Let's also look at the inverse:
Multiply both sides by :
Divide both sides by :
Take the square root of both sides:
.
So can't be 0 and it also can't be positive because the inside of the square root will be negative (since negative divided by positive results in negative).
