Yeah, it would be B (1.2 x 102 m^3) because the measurement gave it away even though other numbers were higher, however, the measurements for those were smaller in size.
Answer:
you didn't ask a question so here is your explanation.
Explanation:
Q = mc∆T. Q = heat energy (Joules, J) m = mass of a substance (kg) c = specific heat (units J/kg∙K) ∆ is a symbol meaning "the change in"
Answer:
Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.
Explanation:
The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:
![Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B%5B%5Cmathrm%7BSO_2%5C%2C%20%28g%29%7D%5D%7D%7B%5B%5Cmathrm%7BO_2%5C%2C%20%28g%29%7D%5D%7D)
.
Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both ![[\mathrm{SO_2\, (g)}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BSO_2%5C%2C%20%28g%29%7D%5D)
and ![[\mathrm{O_2\, (g)}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BO_2%5C%2C%20%28g%29%7D%5D)
will increase if the pressure is increased through compression. However, because 
and 
have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient 
.
As a result, the increase in pressure will have no impact on the value of 
. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.
Answer:
The sugar in DNA is deoxyribose. ... Nucleotides in DNA contain four different nitrogenous bases: Thymine, Cytosine, Adenine, or Guanine. There are two groups of bases: Pyrimidines: Cytosine and Thymine each have a single six-member ring.
It would be considered a Homogeneous Mixture. A mixture with two or more components mixed evenly is a Homogeneous mixture.
