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4 years ago

2. Natural neon exists as a mixture of 90.92% neon-20, 0.26% neon-21, and 8.82% neon-

Chemistry

1 answer:

Paul [167]4 years ago

8 0

Can u please help me on mine

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Which equation best describes the net changes based upon the observation that solid silver nitrate and solid potassium chloride

stealth61 [152]

Answer:

The correct option is: AgNO₃(aq) + KCl(aq) = AgCl(s) + KNO₃(aq)

Explanation:

Precipitation reaction is a chemical reaction that involves reaction between two soluble salts to give an insoluble salt. This insoluble salt exists as a solid and settles down.

Therefore, the solid formed in a precipitation reaction is known as the precipitate.

As the solid silver nitrate (AgNO₃) and solid potassium chloride (KCl) are soluble in water, therefore, their aqueous solutions are represented as AgNO₃(aq) and KCl(aq), respectively.

The precipitation reaction of AgNO₃(aq) and KCl(aq) gives an insoluble salt, silver chloride (AgCl) and a soluble salt, potassium nitrate (KNO₃).

The insoluble salt, AgCl is called the precipitate and is represented as AgCl(s). Whereas, the soluble salt, KNO₃ is represented as KNO₃ (aq).

Therefore, the chemical equation for this precipitation reaction is:

AgNO₃(aq) + KCl(aq) → AgCl(s) + KNO₃(aq)

6 0

3 years ago

In a chemical reaction, __________ are the substances present after the reaction.

chubhunter [2.5K]

Answer:

products

Explanation:

4 0

3 years ago

Read 2 more answers

Why do you think people cannot drink seawater?

Oksanka [162]

Answer:

Its contaminated

Explanation:

5 0

3 years ago

When she introduced CO(g) and Cl2(g) into a 1.00 L evacuated container, so that the initial partial pressure of CO was 1.86 atm

cupoosta [38]

Answer:

Kp is 0.00177

Explanation:

We state the equilibrium:

CO(g) + Cl₂(g) ⇆ COCl₂(g)

Initially we have these partial pressures

1.86 atm for CO and 1.27 for chlorine.

During the reaction, x pressure has been converted. As we have 0.823 atm as final pressure in the equilibrium for COCl₂, pressure at equilibrium for CO and chlorine will be:

1.86 - x for CO and 1.27 - x for Cl₂.

And x is the pressure generated for the product, because initially we don't have anything from it. So pressure in equilibrium for the reactants will be:

1.86 - 0.823 = 1.037 atm for CO

1.27 - 0.823 = 0.447 atm for Cl₂

Let's make, expression for Kp:

Partial pressure in eq. for COCl₂ / P. pressure in eq. for CO . P pressure in eq. for Cl₂

Kp = 0.823 / (1.037 . 0.447) → 0.00177

5 0

3 years ago

10. A 20.00 mL sample of 0.150 mol/L ammonia (NH3(aq)) is titrated to the equivalence point by 20.0 mL of a solution of 0.150 mo

Natalija [7]

Answer:

$\large \boxed{\rm a)\, NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \,$ NH_{4}^{+}(aq) +\text{I}^{-}(aq);\,b)\,11.22;\, c)\, 5.19}$

Explanation:

a) Balanced equation

The balanced chemical equation for the titration is

$\large \boxed{\rm NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \,$ NH_{4}^{+}(aq) +\text{I}^{-}(aq)}$

b) pH at start

For simplicity, let's use B as the symbol for NH₃.

The equation for the equilibrium is

$\rm B + H_{2}O \, \rightleftharpoons\,BH^{+} + OH^{-}$

(i) Calculate [OH]⁻

We can use an ICE table to do the calculation.

B + H₂O ⇌ BH⁺ + OH⁻

I/mol·L⁻¹: 0.150 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.150 - x x x

$K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.150 - x} = 1.8 \times 10^{-5}$

Check for negligibility:

$\dfrac{0.150 }{1.8 \times 10^{-5}} = 8300 > 400\\\\x \ll 0.150$

(ii) Solve for x

$\dfrac{x^{2}}{0.150} = 1.8 \times 10^{-5}\\\\x^{2} = 0.150 \times 1.8 \times 10^{-5}\\x^{2} = 2.7 \times 10^{-6}\\x = \sqrt{2.7 \times 10^{-6}}\\x = \text{[OH]}^{-} = 1.64 \times 10^{-3} \text{ mol/L}$

(iii) Calculate the pH

$\text{pOH} = -\log \text{[OH}^{-}] = -\log(1.64 \times 10^{-3}) = 2.78\\\\\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.78 = \mathbf{11.22}\\\\\text{The pH of the solution at equilibrium is } \large \boxed{\mathbf{11.22}}$

(c) pH at equivalence point

(i) Calculate the moles of each species

$\text{Moles of B} = \text{Moles of HI} = \text{20.00 mL} \times \dfrac{\text{0.0150 mmol}}{\text{1 mL}} = \text{3.00 mmol}$

B + HI ⇌ BH⁺ + I⁻

I/mol: 3.00 3.00 0

C/mol: -3.00 -3.00 +3.00

E/mol/: 0 0 3.00

(ii) Calculate the concentration of BH⁺

At the equivalence point we have a solution containing 3.00 mmol of NH₄I

Volume = 20.00 mL + 20.00 mL = 40.00 mL

$\rm [BH^{+}] = \dfrac{\text{3.00 mmol}}{\text{40.00 mL}} = \text{0.0750 mol/L}$

(iii) Calculate the concentration of hydronium ion

We can use an ICE table to organize the calculations.

BH⁺+ H₂O ⇌ H₃O⁺ + B

I/mol·L⁻¹: 0.0750 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.0750 - x x x

$K_{\text{a}} = \dfrac{K_{\text{w}}} {K_{\text{b}}} = \dfrac{1.00 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\\\\\dfrac{x^{2}}{0.0750 - x} = 5.56 \times 10^{10}\\\\\text{Check for negligibility of }x\\\dfrac{0.0750}{5.56 \times 10^{-10}} = 1.3 \times 10^{6} > 400\\\\\therefore x \text{ $\ll$ 0.0750}$

$\dfrac{x^{2}}{0.0750} = 5.56 \times 10^{-10}\\\\x^{2} = 0.0750 \times 5.56 \times 10^{-10}\\x^{2} = 4.17 \times 10^{-11}\\x = \sqrt{4.17 \times 10^{-11}}\\\rm [H_{3}O^{+}] =x = 6.46 \times 10^{-6}\, mol \cdot L^{-1}$

(iv) Calculate the pH

$\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{6.46 \times 10^{-6}} = \large \boxed{\mathbf{5.19}}$

The titration curve below shows the pH at the beginning and at the equivalence point of the titration.

8 0

3 years ago