A Diatom is under the phylum of Chrysophyta with the following characteristics:
1) Unicellular but often colonial
2) manufactures the carbohydrate chrysolaminarin
3) has unique double -shells of opaline silica
4) contains chlorophylls a and c
They are abundant in sea and fresh water where they serve as producers.
Diatoms are protists that have glass walls. Just like plants, they carry out cytokinesis. They are photosynthesizers with hard shells of silica. An example of a Diatom is Diatoma.
Diatoms are divided into two major groups, one with radial symmetry and the other with bilateral symmetry.
Answer:
3.0 moles.
Explanation:
- It is a stichiometry problem.
- The chemical reaction of reacting hydrogen with oxygen to produce water is:
<em>H₂ + 1/2 O₂ → H₂O.</em>
- It is clear that <em><u>1.0 mole of H₂</u></em> reacts with 0.5 mole of O₂ to produce <u><em>1.0 mole of water</em></u>.
- The ratio of the reacting hydrogen to the produced water is 1:1.
∴ The number of moles of water created from reacting 3.0 moles of hydrogen completely with excess oxygen = 3.0 moles.
<span>The speed of sound in air is about 340 m/s, while the speed of sound in water is about 1,500 m/s. In general, sound travels faster in denser media, so the fish hears the noise first.</span>
Maybe by protons i forgot how to spell it :(
Answer:
The enthalpy of the reaction is coming out to be -380.16 kJ.
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2O%29%7D%29%2B%282%20mol%5Ctimes%5CDelta%20H_f_%7B%28H_2O%29%7D%20%29%5D-%5B%281%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2H_4%29%7D%29%2B%281%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2O_4%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%20mol%5Ctimes%2081.6%20kJ%2Fmol%29%2B2%20mol%5Ctimes%20-241.8%20kJ%2Fmol%29%5D-%5B%281%20mol%5Ctimes%20%2850.6%20kJ%2Fmol%29%29%2B%281%20mol%5Ctimes%20%289.16%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-380.16%20kJ)
Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.
