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2 years ago

The dihydrogenphosphate ion, H2PO4− is amphiprotic. In which of the following reactions is this ion serving as a base?

Chemistry

1 answer:

stira [4]2 years ago

4 0

Answer:

H3PO4(aq) + HPO42-(aq) ⇄ 2H2PO4-

Explanation:

In the reaction of Phosphonic acid, H3PO4 one proton H+ is removed to form dihydrogen phosphate ion, H2PO4-, which is a base.

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How does 0.5 m sucrose (molecular mass 342) solution compare to 0.5 m glucose (molecular mass 180) solution?

mash [69]

Answer : Both solutions contain $3.011 X 10^{23}$ molecules.

Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain $3.011 x 10^{23}$ molecules.

Avogadro's Number is $N_{A}$ = $6.022 X 10^{23}$ which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.

Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.

Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.

Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.

We can calculate the number of molecules for each;

Number of molecules = $N_{A} X M$ ;

∴ Number of molecules = $6.022 X 10^{23} X 0.5 mol/L X 1 L$ which will be = $3.011 X 10^{23}$

Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.

7 0

2 years ago

Read 2 more answers

A mixture of Na2CO3 and MgCO3 of mass 7.63 g is reacted with an excess of hydrochloric acid. The CO2 gas generated occupies a vo

Svetlanka [38]

Answer:

58.6 % by mass of Na₂CO₃

Explanation:

This is the reaction:

Na₂CO₃ + MgCO₃ + 4HCl → MgCl₂ + 2NaCl + 2CO₂ + 2H₂O

Let's find out the moles of CO₂ produced, by the Ideal Gases Law

1.24 atm . 1.67 L = n . 0.082 . 299K

(1.24 atm . 1.67 L / 0.082 . 299K) = n

0.0844 moles = n

Ratio is 2:1, so 2 moles of dioxide were produced by 1 mol of sodium carbonate. Let's make a rule of three:

2 moles of CO₂ were produced by 1 mol of Na₂CO₃

Then, 0.0844 moles of Co₂ would beeen produced by (0.0844 .1)/2 = 0.0422 moles of Na₂CO₃.

Let's convert this moles into mass (mol . molar mass)

0.0422 mol . 106 g/mol = 4.47 g

Finally we can know the mass percent of sodium carbonate in the mixture

(Mass of compound /Total mass) . 100 → (4.47 g / 7.63g) . 100 = 58.6 %

3 0

1 year ago

Rhett is solving the quadratic equation 0= x2 – 2x – 3 using the quadratic formula. Which shows the correct substitution of the

bekas [8.4K]

The correct values I believe would be a=1 b=-2 and c=-3.

8 0

2 years ago

The table shows columns that Brenda uses for her notes on the properties of elements. Her notes state that some elements can rea

skad [1K]

Answer:

<u>in the columns for metals and for metalloids</u>

Explanation:

There are six elements that are always classified as metalloids: boron, silicon, germanium, arsenic, antimony, and tellurium. Pollonium is also, generally, classified as a metalloid,

Metalloids have intermediate electronegativity values (in between that of metals and nonmetals), which is responsible for some similarities (or in between properties) with metals and some similarites with non metals.

An example of such properties that metals and metalloids have in common is that they have relative high melting points. Metalloids are all solid at room temperature, such as most metals.

Other property that both metals and metalloids share is that they can react with oxygen to form oxides that are amphoteric.

Amphoteric compounds are substances that can behave as a base or as an acid, depending on the other compound with which they react.

For instance, among metal oxides, aluminum hydroxide, Al(OH)₃, will act as a base when reacts with hydrochloric acid, HCl, and will react as an acid when reacts with sodium hydroxide, NaOH.

The oxides of metalloids are usually amphoteric.

8 0

2 years ago

Read 2 more answers

Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol

prohojiy [21]

Hey there!:

From the given data ;

Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar = g/L * mol/g

[E]t = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹

Hope that helps!

4 0

2 years ago