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Rashid [163]
2 years ago
12

When 100 ml of 1.0 M Na3PO4 is mixed with 100 ml of 1.0 M AgNO3,a yellow precipitate forms and Ag+ becomes negligibly small. Whi

chof the following is the correct listing of the ions remaining in solutionin order of increasing concentration?(A) PO43- < NO3- < Na+(B) PO43- < Na+ < NO3-(C) NO3- < PO43- < Na+(D) Na+ < NO3- < PO43-(E) Na+ < PO43- < NO3-
Chemistry
2 answers:
ale4655 [162]2 years ago
8 0

Hi! I answered this question once before.

Read more on Brainly.com - brainly.com/question/14596432#readmore

Answer:

A. PO43- < NO3- < Na+

Explanation:

Here’s the reaction equation:

Na3PO4 + 3AgNO3 = Ag3PO4 + 3Na+ + 3NO3-

All of the Ag and PO43- in the solution reacted to form the Ag3PO4 which is the yellow precipitate, leaving little or no Ag or PO43- left in the solution. This narrows down the answer to options A and B.

Since the concentrations and the volumes of the reactants are the same, the number of moles will also be the same.

Na3PO4 will release 3 moles of Na+ while AgNO3 releases 1 mole of NO3-. Hence, the Na+ would be more in solution than the NO3-, narrowing the answer further down to option A.

PO43- < NO3- < Na+

jok3333 [9.3K]2 years ago
4 0

Answer:

Option A

Explanation:

Number of millimoles of Na3PO4 = 1 × 100 = 100

Number of millimoles of AgNO3 = 1 × 100 = 100

When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion

When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-

As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible

Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-

And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-

∴ Increasing order of concentration will be  PO43- < NO3- < Na+

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Answer is: Benzene is trigonal (or triangular) planar.

VSEPR theory (The Valence Shell Electron Pair Repulsion Theory) uses the AXE notation (m and n are integers, m + n = number of regions of electron density).

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m = 3; the number of atoms bonded to the central atom.

n = 0; the number of lone pairs on the central atom.

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1 year ago
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Assuming ideal behavior, how many moles of argon would you need to fill a 14.0×12.0×10.0 ft room? assume atmospheric pressure of
Blababa [14]

We use the formula:

PV = nRT

First let us get the volume V:

volume = 14 ft * 12 ft * 10 ft = 1,680 ft^3

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n = PV / RT

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4 0
1 year ago
1. Concentrated HCl is 11.7M. What is the
SSSSS [86.1K]

Answer:

The concentration is 0,2925M

Explanation:

We use the formula

C initial  x V initial = C final x V final

11,7 M x 25 ml = C final x 1000 ml

C final= (11,7 M x 25 ml)/1000 ml = 0, 2925 M

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6 0
2 years ago
Wine has a pH of 3, which means it is __________ times more acidic than tomatoes, which have a pH of 4.
Natasha2012 [34]

Answer:

10

Explanation:

pH is defined as the negative logarithm of the concentration of hydrogen ions.

Thus,  

pH = - log [H⁺]

Thus, from the formula, more the concentration of the hydrogen ions or more the acidic the solution is, the less is the pH value of the solution.

Thus, solution with pH = 3 will be more acidic than solution with pH =4  

Thus, concentration of the [H⁺] when pH =3

3 = - log [H⁺]

[H⁺] = 10⁻³ M

For pH = 4, [H⁺] = 10⁻⁴ M

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2 years ago
Silver chloride is formed by mixing silver nitrate and barium chloride solutions. What volume of 1.50 M barium chloride solution
konstantin123 [22]

Answer:

1.22 mL

Explanation:

Let's consider the following balanced reaction.

2 AgNO₃ + BaCl₂ ⇄ Ba(NO₃)₂ + 2 AgCl

The molar mass of silver chloride is 143.32 g/mol. The moles corresponding to 0.525 g are:

0.525 g × (1 mol/143.32 g) = 3.66 × 10⁻³ mol

The molar ratio of AgCl to BaCl₂ is 2:1. The moles  of BaCl₂ are 1/2 × 3.66 × 10⁻³ mol = 1.83 × 10⁻³ mol.

The volume of 1.50 M barium chloride containing 1.83 × 10⁻³ moles is:

1.83 × 10⁻³ mol × (1 L/1.50 mol) = 1.22 × 10⁻³ L = 1.22 mL

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