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Roman55 [17]
3 years ago
14

Enter a balanced equation for the reaction between aqueous lead(II)(II) nitrate and aqueous sodium chloride to form solid lead(I

I)(II) chloride and aqueous sodium nitrate. Express your answer as a chemical equation. Identify all of the phases in your answer. Pb(NO3)2(aq)+2NaI(aq)→PbI2(s)+2NaNO3(aq) You have already submitted this answer. Enter a new answer. No credit lost. Try again.
Chemistry
1 answer:
aev [14]3 years ago
3 0

Answer:

Pb(NO₃)₂ (aq)  + 2NaCl (aq)  → PbCl₂ (s) ↓  + 2NaNO₃ (aq)

Explanation:

The reactants are:

Lead(II) nitrate → Pb(NO₃)₂ (aq)

Sodium chloride → NaCl (aq)

The products are:

Lead(II) chloride → PbCl₂ (s)

Sodium nitrate → NaNO₃ (aq)

Salts form nitrate are soluble. The chloride makes a precipitate with the Pb²⁺. The chemical equation for this reaction is:

Pb(NO₃)₂ (aq)  + 2NaCl (aq)  → PbCl₂ (s) ↓  + 2NaNO₃ (aq)

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A HOMOGENEOUS LIQUID THAT CANNOT BE SEPARATED INTO ITS COMPONENTS BY DISTILLATION BUT CAN BE DECOMPOSED BY ELECTROLYSIS IS CLASS
nata0808 [166]

Answer:

ELEMENTS

Explanation:

CUZ AN A

ELEMENT IS A GROUP OF ATOMS THAT CANNOT BE BROKEN DOWN BY ANY CHEMICAL OR PHYSICAL MEAN

7 0
2 years ago
Some amount of hydrogen peroxide (H2O2) breaks down to produce 3 molecules of oxygen (O2) and 6 molecules of water (H2O). How ma
soldier1979 [14.2K]

Answer:

Atoms_H=12Atoms

Explanation:

Hello,

In this case, the only source of hydrogen is in the 6 molecules of water, therefore, the atoms of hydrogen, by applying stoichiometry with the Avogadro's number is:

Atoms_H=6moleculesH_2O*\frac{1molH_2O}{6.022x10^{23}moleculesH_2O}\frac{2molH}{1molH_2O}*\frac{6.022x10^{23}AtomsH}{1molH} \\Atoms_H=12Atoms

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3 0
2 years ago
Read 2 more answers
What is the answer to question 7, NH4C2H3O2
Montano1993 [528]
I think it got deleted but the answer is Ammonium acetate maybe this is why it got deleted 10-9=1
3 0
3 years ago
(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct
Zarrin [17]

Explanation:

The given data is as follows.

   \Lambda^{o}_{m}(NaCl) = 1.264 \times 10^{-2}

   \Lambda^{o}_{m}(H-O=C-ONO) = 1.046 \times 10^{-2}

   \Lambda^{o}_{m}(HCl) = 4.261 \times 10^{-2}

Conductivity of monobasic acid is 5.07 \times 10^{-2} S m^{-1}

     Concentration = 0.01 mol/dm^{3}

Therefore, molar conductivity (\Lambda_{m}) of monobasic acid is calculated as follows.

                 \Lambda_{m} = \frac{conductivity}{concentration}

                                  = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}

                                 = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}

                                 = 5.07 \times 10^{-3} S m^{2} mol^{-1}

Also, \Lambda^{o}_{m} = \Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}

                            = 4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}

                            = 4.043 \times 10^{-2} S m^{2} mol^{-1}

Relation between degree of dissociation and molar conductivity is as follows.

               \alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}

                             = \frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}

                             = 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

Putting the values into the above formula we get the following.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

                        = \frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}

                        = 0.017973 \times 10^{-2}

                       = 1.7973 \times 10^{-4}

Hence, the acid dissociation constant is 1.7973 \times 10^{-4}.

Also, relation between pK_{a} and K_{a} is as follows.

                 pK_{a} = -log K_{a}

                              = -log (1.7973 \times 10^{-4})

                              = 3.7454

Therefore, value of pK_{a} is 3.7454.

                             

3 0
2 years ago
At 25 °C, what is the hydroxide ion concentration, [OH−] , in an aqueous solution with a hydrogen ion concentration of [H+]=4.8×
algol [13]

[OH⁻]= 1 x 10⁻¹⁴ : 4.8 x 10⁻⁴ = 2.083 x 10⁻¹¹

3 0
1 year ago
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