# Tag Info

0

$\ce{CaCO3.MgCO3}$ reacts with $\ce{4 HCl}$, therefore $m_\mathrm{eq}=m/4$. $\ce{CaCO3}$ reacts with $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/2$. $\ce{MgCO3}$ reacts with $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/2$. $\ce{Na2CO3.10 H2O}$ reacts with $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/4$. $\ce{1 CO2}$ of carbonate origin is released by $\ce{2 HCl}$, ...

0

Your method in using the equivalent concept is correct, but the miscorrect in your solution is in determining the n-factor. Refer to the above equations and the calculation N-factorPoutnik N-factor Is the change of atom oxidation state multiplied by the number of atoms changing this state. Because the iodine is just an intermediate product to be titrated ...

2

N-factor Is the change of atom oxidation state multiplied by number of atoms changing this state. Therefore, n-factor of $\ce{I2}$ is $2$, n-factor of $\ce{Na2S2O3}$ is $1$. So, equivalent mass of $\ce{Na2S2O3}$ is equal to its molar mass, for molecular iodine it is half of its molar mass.

1

I can think of another solution, which is somewhat more tedious, but as long as you asked to think differently, here it is: you can utilize the fact that both components are alcohols and share the same empirical formula $\ce{C_yH_{2y+2}O}$, $y\in (1,2)$. This allows to find the unknown amount from the average molar mass $\bar{M}$ of the mix: $$\ce{C_yH_{2y+... 3 Good news is you are in right track and set up two equations correctly. However, I think the calculations would get easier if you have chosen your units of variables in \pu{mol}. Also, you may not needed to do more calculations since the answer is in \pu{mol}. Thus, your two equations should be:$$32x + 46y = 2.16 \;\text{.... (1), and}\frac{3}{2}...

0

To find the molarity $c$, you need to know the amount $n$ required for a titration: $$c(\ce{Na2S2O3}) = \frac{n(\ce{Na2S2O3})}{V(\ce{Na2S2O3})}$$ Unknown amount $n(\ce{Na2S2O3})$ can be found from the second balanced reaction: $$\ce{I2 + 2 Na2S2O3 -> 2 NaI + Na2S4O6}$$ $$n(\ce{Na2S2O3}) = 2\cdot n(\ce{I2})$$ Finding $n(\ce{I2})$ is trivial if you ...

1

Refer to the definition of equivalent (IUPAC Gold Book): equivalent entity Entity corresponding to the transfer of a $\ce{H+}$ ion in a neutralization reaction, of an electron in a redox reaction, or to a magnitude of charge number equal to 1 in ions. In other words, $\pu{1 equiv}$ is the amount of substance reacting with $\pu{1 mol}$ of hydrogen ...

1

If equation $$\ln\frac{[A]_t}{[A]_0}=-2kt$$ were to be used, the answer would ended up being $\pu{33s}$. You cannot substitute $k=2.8\cdot{10^{-8}}$ in the above equation because of the value of $k$ in the last equation of your question equal half the value of k in equation(3.2). Why the values of k in the two equations are different? Because they are ...

2

The key point which you will study in advanced classes, is the concept of elementary reactions. When a given reaction is an elementary reaction, the stoichiometric coefficients become the power of the rate law or in better words, the reaction order corresponds to the stoichiometric coefficients. Definition of elementary reaction: An elementary reaction is a ...

3

What is the sum of the coefficients $(a + b + c)$ in the following reaction? $$\ce{x Zn + y HNO3 -> a Zn(NO3)2 + b H2O + c NH4NO3}$$ This is a redox reaction. During OP's comments, it was clear that he/her got the correct answer, but it wasn't clear that OP understand this is a redox reaction. To balance the equation, it may be better work on with ...

3

Linear equations aside, note that this is a redox reaction and you can use half-reactions method as well (BTW the reaction looks fine to me): $$\ce{x \overset{0}{Zn} + y H\overset{+5}{N}O3 -> a \overset{+2}{Zn}(NO3)2 + b H2O + c \overset{-3}{N}H4NO3}$$ \begin{align} \ce{\overset{+5}{N} + 8 e- &→ \overset{-3}{N}} \tag{red}\\ \ce{\overset{0}{... 1 Different compounds always have to have the same number of each type of atom in them. For example, in respiration:\ce{C6H12O6 + O2 -> CO2 + H2O} The chemical formulas for these compounds must always remain the same else it is no longer that compound. With this in mind balancing increases the number of molecules of each compound you have to ensure ...

3

How is it that a chemical reaction produces an equation, that is essentially an impossibility, according to the law of conservation of mass? You find unbalanced equations as exercises and as a step in writing a balanced equation when there is insufficient information that still needs to be discovered. Balancing equations as an exercise The observable ...

5

First, provided the ball is an atom, you are not writing a chemical equation for a dozen of balls, rather a dozen times at least $10^{16}$ (to speak of equilibrium), usually a dozen times $10^{23}$ (macroscale that normal people operate on). That's not the level you want to use addition for; smaller numbers are easier to operate with. Second, it's not ...

1

I agree with Poutnik's point that this is a poor question. Oxalic acid has two acidic protons so its equivalent weight is $\frac{1}{2}$ its molecular weight. Thus: Answer A is false. Answer B is true. Answer C is false since there are 2*100*0.2 = 40 milliequivalents of acid, but only 100*0.2 = 20 milliequivalents of base. Answer D requires some extra ...

0

The fundamental idea behind gram equivalents and normality is that one gram equivalent of an acid reacts with one gram equivalent of a base. This idea eliminates the need of using molarity and thinking about mole ratios for any acid base titration. Equal volume 1 N HCl will react with 1 N NaOH. Similarly, 1 liter of 1 N $\ce{H2SO4}$ or 1 N $\ce{H3PO4}$ ...

Top 50 recent answers are included