Answer:
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Explanation:
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The minimum mass of NaHCO3 that must be added to the spill to neutralize the acid is 27.216 grams
<h3>calculation </h3>
write the balanced chemical equation
2NaHCO3 +H2SO4 → Na2SO4 +2H2O +2CO2
find the moles of H2SO4 = molarity x volume in liters
volume in liters = 27/1000=0.027 l
moles is therefore= 0.027 x6=0.162 moles
by use of mole ratio of NaHCO3: H2SO4 which is 2:1 the moles of NaHCO3=0.162 x2=0.324 moles
mass of NaHCO3= moles of NaHCO3 x molar mass of NaHCO3(84g/mol)
= 84g/mol x 0.324=27.216 grams
The answer you are looking for would be Colloid, because in colloid mixtures, the particles are big enough to reflect them in a way. Colloids can be distinguished from solutions using the Tyndall effect. Light passing through a colloidal dispersion, such as smoke or foggy air, will be reflected by the larger particles and the light beam will be visible. <span />
Livermorium and tennessine each of these atoms contain 177 neutrons
Answer:
The rate law for second order unimolecular irreversible reaction is
![\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%20k.t%20%2B%20%5Cfrac%7B1%7D%7B%5BA%5D_%7B0%7D%20%7D)
Explanation:
A second order unimolecular irreversible reaction is
2A → B
Thus the rate of the reaction is
![v = -\frac{1}{2}.\frac{d[A]}{dt} = k.[A]^{2}](https://tex.z-dn.net/?f=v%20%3D%20-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3D%20k.%5BA%5D%5E%7B2%7D)
rearranging the ecuation
![-\frac{1}{2}.\frac{k}{dt} = \frac{[A]^{2}}{d[A]}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bk%7D%7Bdt%7D%20%3D%20%5Cfrac%7B%5BA%5D%5E%7B2%7D%7D%7Bd%5BA%5D%7D)
Integrating between times 0 to <em>t </em>and between the concentrations of ![[A]_{0}](https://tex.z-dn.net/?f=%5BA%5D_%7B0%7D)
to <em>[A].</em>
![\int\limits^0_t -\frac{1}{2}.\frac{k}{dt} =\int\limits^A_{0} _A\frac{[A]^{2}}{d[A]}](https://tex.z-dn.net/?f=%5Cint%5Climits%5E0_t%20-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bk%7D%7Bdt%7D%20%3D%5Cint%5Climits%5EA_%7B0%7D%20_A%5Cfrac%7B%5BA%5D%5E%7B2%7D%7D%7Bd%5BA%5D%7D)
Solving the integral
