Answer:
A/1. 10.9 mol O2
Explanation:
583 g x 1 mol SO3 x 3 mol O2 /
80.057 g mol SO3 x 2 mol SO3
- You just need to find molar mass of SO3, which is 80.057 g.
- Everything else came from formula. Further explanation...
- Always start with what they give, such as 583 g. Then find 1 mol of what is being produced, in this it is SO3. We already found this because we did molar mass above. Next. find how many moles of what they want, which is O2. Look in equation and you can see 3 mol in from of O2. Next, do the same for SO3 and you can find 3 mol in front of that. Lastly, just do the math.
- If you need a further explanation or more help on any problems I would be happy to help, just let me know.
Ionic bonds form when a nonmetal and a metal exchange electrons, while covalent bonds form when electrons are shared between two nonmetals. An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions.
Answer:
The correct answer is "Fragment B likely has a higher Guanosine/Citosine content".
Explanation:
Guanosine/Citosine content, or GC content, refers to how many molecules of guanosine and citosine have a DNA fragment, respect to the content of adenine and thymine. The higher the GC content, the higher the temperature needed to denature the fragment of DNA. This happens because guanosine and citosine establish three hydrogen bonds, while adenine and thymine establish two hydrogen bonds when they bind together. Therefore, if fragment A and B are the same length, but at 89 C only fragment A is completely denatured, fragment B likely has a higher GC content.
Answer:
The concentration of the pyridinium cation at equilibrium is 1.00×10⁻³ M
Explanation:
In water we have
C₅H₅NHBr + H₂O ⇒ C₅H₅NH+ + Br−
Pyridinium Bromide (C₅H₅NHBr) Dissociates Completely Into C₅H₅NH+ And Br− as such it is a strong Electrolyte.
Therefore the number of moles of positive ion produced per mole of C₅H₅NHBr is one
pH = - log [H₃O⁺] Therefore 10^-pH = [H₃O⁺] = concentration of C₅H₅NHBr
= 10⁻³ = 0.001M = concentration of C₅H₅NHBr
The concentration of C₅H₅NHBr is = 1.00×10⁻³ M to two places of decimal
