10

Plants are able to store energy as carbohydrates, i.e. they can make more carbohydrates than they need for their metabolism when there is no sunlight, and store it for later use. So the amount of energy in your equations is not equal. Some plants are offering this stored energy as sugars as nectar or fruits in exchange for pollination and spreading their ...


7

Biomass The net production of $\ce{CO2}$ in all living organisms is close to zero. The argument is that if the total biomass on earth does not change over a period of time, and the net transfer from biomass to other reservoirs of carbon does not change much, the production and consumption should balance out. Wikipedia's article on biomass states: The total ...


6

I am afraid you are mixing all modern and old concepts. Moles and the concept of limiting reagents did not exist in Richter's time. It took 615 parts by weight of magnesia (MgO), for example, to neutralize 1000 parts by weight of sulfuric acid This relatively famous statement has nothing to do with law of multiple proportions but it illustrates the ...


5

If by "fancier" you mean "uglier", then absolutely yes. Real chemistry is messy, much more than we can fully handle. The simple rate laws we learn in undergraduate physical chemistry are just approximations. I've written elsewhere that, in general, a full description of chemical kinetics for any overall reaction requires studying an entire tree of ...


4

Sometimes, practical chemistry does not work like it's supposed to on paper. A perfectly balanced reaction equation may dictate that two reactants should be in a 1:1 stoichiometric ratio, but in practice, you get much better yields or reaction times if you use an excess of one. When that happens, you are no longer using a stoichiometric amount, as specified ...


4

There is very well-written chapter "The chemistry of monovalent copper in aqueous solutions" in Advances in inorganic chemistry, Volume 64 [1, pp. 220–223, DOI: 10.1016/B978-0-12-396462-5.00007-6] which extensively covers as to why copper(I) is unlikely to exist in aqueous solution and why nitrate is a poor ligand for the purpose of preserving monovalent ...


4

Generally, other than the seven diatomic gases at STP ($\ce{H2}$, $\ce{O2}$, $\ce{N2}$, $\ce{F2}$, $\ce{Cl2}$, $\ce{Br2}$, $\ce{I2}$... though $\ce{Br2}$ & $\ce{I2}$ have fairly low vapor pressure at room temperature), for convenience, other elements are considered as if they reacted as individual atoms, though, as you state, that is not actually the ...


4

Your second equation (respiration) happens less than your first equation (photosynthesis) while the plant is growing. About a third of the mass of a typical plant is cellulose, which is created by linking the glucose that is product of the first reaction. So the fact that you have a plant in front of you at all means that the plant has performed the first ...


4

Fortunately this issue was just caused by a small oversight. In the midst of so many coefficients, you accidentally skipped the second nitrogen atom in $\ce{NH4NO3}$, which changes the fifth element in the fourth row from 1 to 2. The correct composition matrix is then: $$\begin{bmatrix} 3 & 8 & 1 & 12 & 4 & 2 \\ 1 & 0 & 0 & 1 ...


3

Eli, you should have read the paper very carefully before asking questions. I think you are very intelligent person, so don't choose easy way out. Now, to answer your question, about mol%. It is right in the paper. As you already figured out and also Orthocresol pointed out in his comment, The amount equals to given mol% is $LR \times \frac{given \ mol\%}{...


3

As noted in the comments, $\ce{Cu+}$ is not stable in aqueous solution. It tends to disproportionate to $\ce{Cu^0}$ and $\ce{Cu^{2+}}$. You can therefore assume that all but a negligible amount turned into $\ce{Cu^{2+}}$. Source: Holleman/Wiberg: Lehrbuch der Anorganischen Chemie, de Gruyter, 101. edition, 1995.


3

Citric acid has $\mathrm{p}K_\mathrm{a}$-values of 3.1, 4.7, and 6.4, while those of tartaric acid are 3.0 and 4.3. If you adjust the $\mathrm{pH}$ to 6.4, tartaric acid would be roughly 100% deprotonated, while 50% of citric acid still has one proton to give off. If you titrate this solution with $\ce{NaOH},$ you could estimate the buffer capacity, giving ...


3

The $K_\mathrm{sp}$ of $\ce{Fe(OH)3}$ is varied from source to source, but a reliable source gives the value of $2.79 \times 10^{-39}$ at $\pu{25 ^\circ C}$. We'll use this value throughout the calculations. Suppose, $s$ amount of $\ce{Fe(OH)3}$ dissolves some in water according to its $K_\mathrm{sp}$, but assume water is not ionized: $$\ce{Fe(OH)3(s) <=&...


3

No, you should not do the problem your way. Given $$(0.08−x)(0.09−x)=10^{−14}$$ You can solve for $x$ here, but its value will be quite close to $0.08$. This is the only way to make the product a small number, by making the one of the multiplicands very small. However, $x$ will be so close to 0.08 that will have trouble figuring out the value of $0.08 - x$, ...


2

Let’s take the simple example of the combustion of hydrogen $\ce{H2}$ in oxygen $\ce{O2}$. The equation of the reaction is: $$\ce{2 H2 + O2 -> 2 H2O}$$ It means that $2$ moles of $\ce{H2}$ $(2\cdot2\ \mathrm g$) react with $1$ mole ($32\ \mathrm g$) of $\ce{O2}$. These masses are in a ratio $4:32=1:8$. All mixtures containing $\ce{H2}$ and $\ce{O2}$ in ...


2

The original equation cannot be solved using mathematics because it gives the coefficient of $\ce{XeO3}$ zero. The reason for that is exact reason given by Ivan Neretin (see else where). Yet, I'd like to show a way of solving a balancing problem of true chemical equation by using mathematics. First, assume the compound is $\ce{XeF4}$, which undergoes ...


2

There is one particle of dsDNA in the cell. Divide by $N_\mathrm{A}$ to get the amount of substance, and divide by the volume to get the concentration matching the concentration given in the answer. \begin{align} V_{\mathrm{cell}} &= 4 \cdot {(\pu{0.5 μm})}^{3}\\[0.5ex] &= \pu{0.5 μm3}\\[0.5ex] &= \pu{5E-16 L}\\[3.5ex] N_{\mathrm{...


2

Equivalent concept is an archaic unit of measurement that was used in chemistry and the biological sciences in the era before researchers knew how to determine the chemical formula for a compound. An equivalent (symbol: officially equiv; unofficially but often Eq) is the amount of a substance that reacts with (or is equivalent to) an arbitrary amount of ...


2

Here we have an already balanced equation given in the question- $$\ce{2 MnI2 + 13 F2⟶ 2 MnF3 + 4 IF5}$$ We find the given number of moles of each reactant by dividing their given masses by their respective molar masses Given mass of $\ce{MnI2}=\pu{1.23g}$ Molar mass of $\ce{MnI2}= \pu{308.74 g/mol}$ Amount of $\ce{MnI2} = \frac{1.23}{308.74}= \pu{0....


2

Sulfur is made of $S_8$ molecules when cooled down from gaseous or liquid state. But when it is obtained by a chemical reaction at room temperature, it is a mixture of many allotropes, like $S_n$, where n is not well defined and may vary. Also, by cooling abruptly liquid sulfur in carbon disulfide and extracting the soluble part, the dissolved sulfur is $...


2

Assuming the water doesn't displace much volume in the bag, the volume a mole of CO2 displaces at room temperatre is 22 L. pV=nRT p=1 atm T=298 K R=0.0821 $\frac{L*atm}{mol K}$ Step 1: find out how much volume the bag is in liters (units need to be consistent) Step 2: Using pV=nRT solve for the number of moles using that volume of CO2 that you need ...


2

We cannot say how many significant figures you should be reporting because: We cannot be sure about the precision of your volume measurements; and We cannot be sure about the precision of the original solution’s concentration. Concerning volumes: most of the time when working with such volumes in the lab, I would use measuring cylinders. These come with a ...


2

As you are new to this forum I want to welcome you but nevertheless I would kindly ask you to first make your own research on a problem you have and then ask your question providing some information on what you do not understand. Your question seems to be a homework which you only copied without any signs of your own thoughts. That said, here are some ...


2

As stated by Zhe, it is difficult to solve the equation as the value of x would be very close to $0.08$. However there is a workaround, if you wish to solve it this way using $K_\mathrm{w}$ from the side of the $\ce{H+}$ ion. The equation that you have is: $$\tag{1} (0.08-x)(0.09-x)=10^{-14}$$ Instead of solving for $x$ here, we rather say $Y = 0.08-x$. Here ...


2

This is an example of a classical method of determining chemical formulas by elemental analysis. You are not supposed to know the chemical formula of copper oxide neither they are asking you to write a balanced equation. All they are telling you is that (i) Mass of pure oxide of copper = 3.45 g (ii) Mass of copper obtained after reduction (we do not need to ...


2

Hydrazine has more hydrogen atoms per mL than water Some simple calculations give the moles of hydrogen per mL in some possible alternative liquids. Taking into account density and molecular mass we get the following results: Water 0.056 mol/mL 2 Hydrogens -> 0.11 mol H/mL Hydrazine 0.0313 mol/mL 4 hydrogens -> 0.125 mol H/mL symmetric ...


2

Let's get the simple question out of the way first: How do you balance a redox reaction lacking a molecular reactant on the product side? I am referring to the atom or molecule that changes oxidation state when I say molecular reactant. You can not balance it then. Also, you wouldn't be given such a reaction and is always a typo when they've said that it ...


2

Dichromate (orange) is present in acidic medium while chromate (yellow) prevails in basic medium. Here are stepwise solutions in both media. Acidic Medium: Recognizing that chromium is the oxidant, balance the chromium atoms. $$\ce{Cr2O7^2- -> 2Cr^3+}$$ Because the reaction is conducted in aqueous acid, only water and protons are available to balance the ...


2

The excess of $\ce{O2}$ is ${7.32~ mol}$. So the excess air is : ${7.32 ~mol· (1 + 3.76) = 34.84~ mol}$. The total amount of air is : ${a·4.76 = 19.53~ mol ·4.76 = 92.96~ mol}$. The percentage of excess air is : $34.84/92.96 = 0.3748 = 37.48$%


2

When a sample of $\ce{CuSO4⋅5H2O}$ is dissolved in water, the $5$ molecules water are mixed with the water used as solvant, and cannot be distinguished any more from the other molecules of water. It does not really matter, as if you want to calculate the molarity, you never dissolve your sample of $\ce{CuSO4·5H2O}$ in exactly $1$ litre water. No ! You ...


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