1000 mL=1L
25 mL = 0.025 L
125 mL = 0.125 L
M1V1=M2V2
0.15(0.125) = M2(0.025)
0.01875 = M2(0.025)
0.75 = M2
0.75 M
Answer:
5 moles of Fe
Explanation:
The equation of the reaction is;
2 Al(s) + Fe 2O 3(s) --> 2Fe (s) + Al 2O 3 (s)
Now;
1 mole of Fe2O3 require 2 moles of Al
3 moles of Fe2O3 requires 3 × 2/1 = 6 moles of Al
Hence Al is the limiting reactant.
If 2 moles of Al yields 2 moles of Fe
5 moles of Al yields 5 × 2/2 = 5 moles of Fe
Number one is physical change the other one is chemical change
Agar is used to assist establish an anaerobic environment that promotes nitrate reduction.
Nitrate Reduction test:
- The nitrate in the broth is converted to nitrite by organisms that can produce the nitrate reductase enzyme, which can then be further converted to nitric oxide, nitrous oxide, or nitrogen.
- Anaerobic respiration and denitrification are two processes that can convert nitrate to a variety of compounds.
- While denitrification only reduces nitrate to molecular nitrogen, anaerobic respiration employs nitrate as the bacterium's final electron acceptor, reducing it to a range of chemicals.
- The nitrate reduction test is based on the detection of nitrite and its capacity to produce a red precipitate (prontosil), which is a water-soluble azo dye, when it combines with sulfanilic acid to create a complex (nitrite-sulfanilic acid).
Learn more about the Nitrate reduction test with the help of the given link:
brainly.com/question/11181586
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This is a dilution that requires a certain volume from the stock solution to be diluted with distilled water to make a solution of HBr with a lesser concentration than the stock solution
Following dilution formula can be used
c1v1 = c2v2
Where c1 is concentration and v1 is the volume of the stock solution
c2 is concentration and v2 is volume of the diluted solution to be prepared
Substituting these values
10.0 M x v1 = 3.0 x 450.0 mL
v1 = 135.0 mL
A volume of 135.0 mL from HBr stock solution needs to be taken and diluted with distilled water upto 450.0 mL. The resulting solution will have a concentration of 3.0 M