The atomic number of an atom refers to its ___

Sort The following iron compound by whether the cation is iron(II) or iron(III)

lora16 [44]

Answer:

Left, Left, Right, Left, Right, Right

Explanation:

Follow that order!

5 0

4 years ago

What monomer is made up of a base, a sugar, and a phosphate group?

balandron [24]

The answer is the option B. Nucleotide. Nucleotides are organic molecules formed by three units :one or more phosphate groups, a sugar with five carbons (ribose or deoxyribose) and a nitrogenous base. Nucleotids are the units that form the two nucleic acid polymers: DNA and the RNA.

6 0

4 years ago

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The standard free-energy changes for the reactions below are given.Phosphocreatine → creatine + Pi ∆ G'° = –43.0 kJ/molATP → ADP

Anton [14]

Answer:

Gibbs free-energy of the reaction = (–12.5 kJ/mol)

Explanation:

The Gibbs free-energy of a reaction predicts the spontaneity or feasibility of a given chemical reaction.

Given the standard Gibbs free energy changes:

Phosphocreatine → creatine + Pi, ∆G° = –43.0 kJ/mol ...(1)

ATP → ADP + Pi , ∆G° = –30.5 kJ/mol ....(2)

Now to calculate the Gibbs free-energy of the given chemical reaction: Phosphocreatine + ADP → creatine + ATP; the equation (2) is reversed to give:

ADP + Pi → ATP, ∆G° = + 30.5 kJ/mol ...(3)

Now the equation (3) and (1) are added, to give:

Phosphocreatine + ADP + Pi→ creatine + ATP + Pi

⇒ Phosphocreatine + ADP → creatine + ATP

Therefore, to calculate the Gibbs free-energy of the reaction, the standard Gibbs free energy changes of the equations (1) and (3) are added similarly:

Gibbs free-energy of the reaction: ∆G° = (–43.0 kJ/mol) + ( + 30.5 kJ/mol) = (–12.5 kJ/mol)

Therefore, the Gibbs free-energy of the reaction = (–12.5 kJ/mol)

7 0

3 years ago

Equilibrium is characterized by

Hatshy [7]

Answer:

the thenhow to be my favorite you want to go don't think I love you too I don't think so ako wala load it out and then the words of the is being proud of you Fouck you Fouck

Explanation:

good morning I love you too baby girl women it's a

6 0

3 years ago

What is the pOH of a 2.6 x 10-6 M H+ solution?

melomori [17]

Answer:

Approximately $8.41$ (assuming that the solution is at $\rm 25^\circ C$ , under which $K_{\rm w} = 10^{-14}$ .)

Explanation:

Let ${\rm [H^{+}]}$ and ${\rm [OH^{-}]}$ denote the concentration of $\rm H^{+}$ and $\rm OH^{-}$ respectively.

Let $K_{\rm w}$ denote the self-ionization constant of water. The exact value of $K_{\rm w}\!$ depends on the temperature of the solution. $K_{\rm w} =10^{-14}$ at $\rm 25^\circ C$ .

The product of ${\rm [H^{+}]}$ and ${\rm [OH^{-}]}$ in a solution (with $\rm M$ , or moles per liter, as the unit) is supposed to be equal to the $K_{\rm w}$ value of that solution at the corresponding temperature. In other words:

${\rm [H^{+}]} \cdot {\rm [OH^{-}]} = K_{\rm w}$ .

Rearrange to obtain an expression for ${[\rm OH^{-}]}$ :

$\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]}\end{aligned}$ .

Assume that the solution in this question is at $\rm 25^\circ C$ (for which $K_{\rm w} =10^{-14}$ .) For this solution:

$\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]} \\ &= \frac{10^{-14}}{2.6 \times 10^{-6}}\approx 3.85\times 10^{-9}\; \rm M\end{aligned}$ .

Hence, the $\rm pOH$ of this solution would be:

$\begin{aligned}\rm pOH &= -\log_{10}{\rm [OH^{-}]} \\&\approx -\log_{10} (3.85 \times 10^{-9}) \approx 8.41 \end{aligned}$ .

3 0

3 years ago

The atomic number of an atom refers to its ____.