Answer: hehe its kirishima here and the various types of fossil fuels are energy provision coal, oil and natural gas. Coal is a solid fossil fuel formed over millions of years by decay of land vegetation.
haha hope this helped
-yours truly red riot
I think it would be better control of fusion reactions.
A wet-chemistry biochemical analyzer<span> was assessed for in-practice veterinary use. Its small size may mean a cost-effective method for low-throughput in-house biochemical analyses for first-opinion practice. The objectives of our study were to determine imprecision, total observed error, and acceptability of the </span>analyzer<span> for measurement of common canine and feline </span>serum<span> analytes, and to compare clinical </span>sample<span> results to those from a commercial reference </span>analyzer<span>. Imprecision was determined by within- and between-run repeatability for canine and feline pooled </span>samples<span>, and manufacturer-supplied quality control material (QCM). Total observed error (TEobs) was determined for pooled </span>samples<span> and QCM. Performance was assessed for canine and feline pooled </span>samples<span> by sigma metric determination. Agreement and errors between the in-practice and reference </span>analyzers<span> were determined for canine and feline clinical </span>samples<span> by Bland-Altman and Deming regression analyses. Within- and between-run precision was high for most analytes, and TEobs(%) was mostly lower than total allowable error. Performance based on sigma metrics was good (σ > 4) for many analytes and marginal (σ > 3) for most of the remainder. Correlation between the </span>analyzers<span> was very high for most canine analytes and high for most feline analytes. Between-</span>analyzer<span> bias was generally attributed to high constant error. The in-practice </span>analyzer<span> showed good overall performance, with only calcium and phosphate analyses identified as significantly problematic. Agreement for most analytes was insufficient for transposition of reference intervals, and we recommend that in-practice-specific reference intervals be established in the laboratory.</span>
Answer:
The rate of the reaction will increase by a factor of 9.
Explanation:
Hello,
In this case, considering the given second-order reaction, whose rate law results:
![r=k[A] [B]^2](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%20%5BB%5D%5E2)
We easily infer that at constant concentration of A but tripling the concentration of B, we are going to obtain the following increasing factor while holding the remaining variables constant:
![Increase\ factor=\frac{r_{final}}{r_{initial}} =\frac{k[A][3*B]^2}{k[A][B]^2} =\frac{3^2}{1} \\Increase\ factor=9](https://tex.z-dn.net/?f=Increase%5C%20factor%3D%5Cfrac%7Br_%7Bfinal%7D%7D%7Br_%7Binitial%7D%7D%20%3D%5Cfrac%7Bk%5BA%5D%5B3%2AB%5D%5E2%7D%7Bk%5BA%5D%5BB%5D%5E2%7D%20%3D%5Cfrac%7B3%5E2%7D%7B1%7D%20%5C%5CIncrease%5C%20factor%3D9)
Best regards.
Answer:
Percentage yield = 30%
Explanation:
Given data:
Number of moles of NO = 7.0 mol
Number of moles of O₂ = 5 mol
Number of moles of NO₂ = 3 mol
Percentage yield = ?
Solution:
Chemical equation:
2NO + O₂ → 2NO₂
Now we will compare the moles of NO₂ with NO and O₂ .
NO : NO₂
2 : 2
7.0 : 7.0
O₂ : NO₂
1 : 2
5.0 : 2 ×5.0 = 10 mol
The number of moles of NO₂ produced by NO are less it will be limiting reactant.
Mass of NO₂ = moles × molar mass
Mass of NO₂ = 10 mol × 46g/mol
Mass of NO₂ = 460 g
Actual yield of NO₂:
Mass of NO₂ = moles × molar mass
Mass of NO₂ = 3 mol × 46g/mol
Mass of NO₂ = 138 g
Percentage yield:
Percentage yield = Actual yield/theoretical yield × 100
Percentage yield = 138 g/ 460 g × 100
Percentage yield = 30%
