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Dafna11 [192]
3 years ago
12

Joanna is performing a reaction that generates a moderate amount of hydroxide. To mimic biological conditions (most bodily fluid

s in living organisms retain a relatively constant pH around 7), she knows that she needs to add a buffer to the reaction. Which of the following will MOST effectively neutralize the OH− ions produced by the reaction
Chemistry
1 answer:
Nady [450]3 years ago
7 0

Answer:

As for your question, I know to forget to put the options, specifically that your question is incomplete.

Explanation:

Although it could help you by telling you that always a reaction that seeks to balance the pH, and achieve neutrality ... It is necessary to achieve a concentration of OH equal to that of H +, in this way the hydroxyl and the protons.

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Una muestra de 1 gramo de un elemento contiene 1,5 x 1022 atomos, cual es su masa molar❓❓
marusya05 [52]

El número de Avogadro es 6,022 x 10^23, y es el número de átomos que hay en un mol de dicho elemento. A su vez un mol es la cantidad de un elemento cuya masa en gramos coincide con el peso atómico.

Por tanto, 6,022 x 10^23 átomos del elemento tienen una masa en gramos igual a su peso atómico. Hacemos una regla de tres:


1 gramo -------- 1,5 x 10^22 átomos

x ------------------ 6,022 x 10^23 átomos


=> x = 40,1 gramos por mol del elemento.


De modo que su peso atómico es 40,1. Se trata del calcio.


Saludos.

7 0
3 years ago
What is the percentage composition of cobalt in cobalt(II) fluoride, CoF2?
emmainna [20.7K]

Answer:

60.8%

Explanation:

We'll begin obtaining the molar mass of cobalt(II) fluoride, CoF2. This can be done as shown below:

Molar mass of CoF2 = 59 + (19x2) = 97g/mol.

The percentage composition of cobalt in cobalt(II) fluoride, CoF2 is given by:

Mass of Co/Molar Mass of CoF2 x 100

=> 59/97 x 100 = 60.8%

Therefore, the percentage composition of cobalt in cobalt(II) fluoride, CoF2 is 60.8%

5 0
2 years ago
How many moles are in 5.60 x 10^23 atoms of Na? How many moles of H2 can be formed from this amount of Na?
alisha [4.7K]

Answer:

\large \boxed{\text{A. 0.930 mol Na; B. 0.465 mol of H}_{2}}

Explanation:

A. Moles of Na

1 mol Na atoms = 5.60 × 10²³ atoms of Na

\text{Moles of Na } = 5.60 \times 10^{23} \text{ atoms Na} \times \dfrac{\text{1 mol Na}}{6.022 \times 10^{23}\text{ atoms Na}}\\\\= \text{0.930 mol Na}\\\text{There are $\large \boxed{\textbf{0.930 mol Na}}$ in 5.60 $\times 10^{23}$ atoms of Na.}

B. Moles of hydrogen

You haven't given  the equation for the reaction. However, we can write an equation

                 2Na + 2HX ⟶ 2NaX+ H₂

n/mol:  0.930  

1 mol of H₂ is formed from 2 mol of Na

\text{Moles of H}_{2} = \text{0.930 mol Na} \times \dfrac{\text{1 mol H}_{2}}{\text{2 mol Na}} = \textbf{0.465 mol H}_{2}\\\\\text{You can form $\large\boxed{\textbf{0.465 mol of H}_{2}}$}

5 0
3 years ago
These images of the Santa Cruz River in Arizona were taken in 1942 and 1989. During the period between these years, the growth o
icang [17]

The roots could no longer access depleted groundwater.


The topsoil in the area eroded.



7 0
3 years ago
6. XY₂ is the molecular formula of an ionic compound, where X is a metal! and also a representative element. Identify the groups
Rufina [12.5K]

XY₂ is the molecular formula of an ionic compound where X is magnesium (Mg) and Y is chloride (Cl) and magnesium is metal of second group of the periodic table and represents the 2nd group and chloride belongs to 17th group.

<h3>What are ionic compound?</h3>

Ionic compounds are those which completely dissociate into their constituent ions in water one is metal and other is non metal.

They complete their octave by sharing lone pair of electron and attend the positive and negative charge after dissociation are know as ionic compounds.

Therefore, the molecular formula of an ionic compound where X is magnesium (Mg) and Y is chloride (Cl) and magnesium is metal of second group of the periodic table and represents the 2nd group and chloride belongs to 17th group.

Learn more about ionic compound, here:

brainly.com/question/7655594

#SPJ1

5 0
1 year ago
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