Answer:
The use of pomace for animal feed might be chosen if minimizing production costs is desired
Explanation:
i've taken the test
Determine the mass in grams of each element in the sample. If you are given percent composition, you can directly convert the percentage of each element to grams.
For example, a molecule has a molecular weight of 180.18 g/mol. It is found to contain 40.00% carbon, 6.72% hydrogen and 53.28% oxygen.
Convert the percentages to grams.
40.00 grams of carbon
6.72 grams of hydrogen
53.28 grams of oxygen
Answer:
2,2,3,3-tetrapropyloxirane
Explanation:
In this case, we have to know first the alkene that will react with the peroxyacid. So:
<u>What do we know about the unknown alkene? </u>
We know the product of the ozonolysis reaction (see figure 1). This reaction is an <u>oxidative rupture reaction</u>. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If 
is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.
<u>What is the product with the peroxyacid?</u>
This compound in the presence of alkenes will produce <u>peroxides.</u> Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product <u>2,2,3,3-tetrapropyloxirane.</u> (see figure 2)
Explanation:
Metals and non-metals can be identified either by their position in the periodic table or by their properties.
he metal elements are found on the left hand side of the periodic table, and the non-metal elements are found on the right. You can imagine a zig-zag line, starting at B-Al-Si, separating metals from non-metals.
Answer:
(1) order = 2
(2) R = K [A]²
Explanation:
Given the reaction:
A--------->Product
The rate constant relation for the reaction is given as:
R(i) = K [A]............(*)
Where R(I) is rate constant at different concentration of A.
Taking the rate constant as R1, R2 and R3 for the different concentrations respectively. Then the following equations results
0.011 = K [0.15] ⁿ.........(1)
0.044 = K [0.30]ⁿ .......(2)
0.177 = K [0.60]ⁿ .........(3)
Dividing (2) by (1) and (3) by (1)
Gives:
0.044/0.011 = [0.3/0.15]ⁿ
4 = 2ⁿ; 2² = 2ⁿ; n = 2
Similarly
0.177/0.011 = [0.60/0.15]ⁿ
16.09 = 4ⁿ
16.09 = 16 (approximately)
4² = 4ⁿ ; n = 2
Hence the order of the reaction is 2.
The rate law is R = K [A]²
