The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]
<h3>What is Balanced Chemical Equation ?</h3>
The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
The equation for aqueous solution of H₂CO₃ is
H₂CO₃ → H₂O + CO₂
The charge balance equation is
[HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]
Thus from the above conclusion we can say that The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]
Learn more about the Balanced Chemical equation here: brainly.com/question/26694427
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Zeff = Z - S
Here, Z is the number of protons in the nucleus, that is, atomic number, and S is the number of nonvalence electrons.
For boron, the electronic configuration is 1s₂ 2s₂ 2p₄
Z = 5, S = 2
Zeff = 5-2 = +3
For O, electronic configuration is 1s₂ 2s₂ 2p₄
Z = 8, S = 2
Zeff = 8-2 = +6
Hence, the correct answer is second option, that is, +3 and +6, the Zeff of boron is smaller in comparison to O, thus, boron exhibits a bigger size than O.
<span>a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.
</span>and % will be 60%.
1 is seconds and meters m/s
2 is seconds and meters m/s^2
3 Newton kg/m
4 Kilograms
Hope this helps!
Answer:
A.
Explanation:
Using the ideal gas equation, we can calculate the number of moles present. I.e
PV = nRT
Since all the parameters are equal for both gases, we can simply deduce that both has the same number of moles of gases.
The relationship between the mass of each sample and the number of moles can be seen in the relation below :
mass in grammes = molar mass in g/mol × number of moles.
Now , we have established that both have the same number of moles. For them to have the same mass, they must have the same molar masses which is not possible.
Hence option A is wrong
