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statuscvo [17]
3 years ago
10

Enter the net ionic equation for the reaction of AgNO3(aq)

Chemistry
1 answer:
mixas84 [53]3 years ago
6 0

Answer:

Explanation:

This is a typical double replacement or double decomposition reaction. In this kind of reaction, it involves the combination of ionic compounds which exchange their partners to form new compounds. Typical of these reaction is the formation of precipitates which are insoluble compounds.

The reaction equation is shown below:

 AgNO₃(_{aq} ) + K₂SO₄(_{aq} ) → 2KNO₃(_{aq} ) + AgSO₄(_{s} )

From the rule of solubility:

  • All trioxonitrate(v) salts are soluble
  • Salts of silver sulfates are slightly soluble. This will form the precipitate in the solution.

We can write the ionic equation as:

      Ag⁺(_{aq} ) + NO₃⁻(_{aq} ) + 2K⁺(_{aq} ) + SO₄²⁻(_{aq} ) → AgSO₄(_{s} ) + 2K⁺ + NO₃⁻(_{aq} )

The spectator ions or non-reactive ions are those ions that appears on both sides of the equation. These ions are the potassium and trioxonitrate ions. We cancel them out to give the net ionic equation of the reaction:

 Ag⁺(_{aq} ) + SO₄²⁻(_{aq} ) → AgSO₄(_{s} )

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Hydrofluoric acid (HF) can be prepared according to the following equation:
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Given the balanced equation:

( Reaction type : double replacement)

CaF2 + H2SO4 → CaSO4 + 2HFI

We can determine the number of grams prepared from the quantity of 75.0 H2SO4, and 63.0g of CaF2 by converting these grams to moles per substance.

This can be done by evaluating the atomic mass of each element of the substance, and totaling it to find the molecular mass.

For H2SO4 or hydrogen sulfate it's molecular mass is the sum of the quantity of atomic mass per element. H×2 + S×1 + O×4 = ≈1.01×2 + ≈32.06×1 + ≈16×4 = 2.02 + 32.06 + 64 = 98.08 u (Dalton's or Da) or g / mol.

For CaF2 or calcium fluoride, it's molecular mass adds 1 atomic mass of calcium and 2 atomic masses of fluoride due to the number of atoms.

Ca×1 + F×2 = ≈40.07×1 + ≈19×2 = 40.08 + 38 = 78.07 u (Da or Dalton's) or g / mol.

3 0
3 years ago
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What greek letter are used in the precise definition of a limit?
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Epsilon-delta would be the green letter that is used in the precision definition of a limit. (ε, δ).
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3 years ago
Chris the chemist was working in the and he wanted to sou what would happen he put copper chloride into a Bunsen flameAs soon as
Reptile [31]

The question is incomplete, the complete question is;

Chris the Chemist was working in the lab, and he wanted to see what would happen if he put copper chloride into a Bunsen burner flame. As soon as he put it in the flame, it began to emit a green color. Which of these BEST describes this outcome?

A) The heat split the nucleus and created a new element.

B) As the copper chloride was heated, electrons were turned into neutrons.

C) As the electrons moved to a higher energy level, a photon of light is emitted.

D) When excited electrons return back to the ground state, a photon of light is emitted.

Answer:

D) When excited electrons return back to the ground state, a photon of light is emitted.

Explanation:

According to Bohr's model of the atom, electrons may absorb energy and move up to higher energy levels.

These electrons quickly return to ground state from such higher energy excited states thereby emitting the excess energy absorbed during excitation.

If the higher energy level is E2 and the ground state is E1 then the energy of the photon emitted when the electron descends from energy level E2 to E1 is;

ΔE = E2 - E1

This photon emitted accounts for the green colour of the copper salt observed when it was heated.

4 0
3 years ago
Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying
PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (K_a for HF is 6.8\times 10^{-4}.)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = K_a=6.8\times 10^{-4}

HF\rightleftharpoons H^++F^-

Initially

c          0            0

At equilibrium :

(c-x)      x             x

The expression of disassociation constant is given as:

K_a=\frac{[H^+][F^-]}{[HF]}

K_a=\frac{x\times x}{(c-x)}

6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

[H^+]=x=0.01346 M

The pH of the solution is ;

pH=-\log[H^+]=-\log[0.01346 M]=1.87

The pH of an 0.280 M HF solution is 1.87.

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3 years ago
How many moles are in 4.5 g of<br> Sodium Chloride, NaCl?
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Answer:

The answer to the question is 0.07 moles

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