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Crazy boy [7]
2 years ago
14

What mass of hydrogen is produced when 192 g of magnesium is reacted with hydrochloric acid?​

Chemistry
2 answers:
Leona [35]2 years ago
5 0

\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}

\longrightarrow \sf{Mg+2HCI \Longrightarrow MgCl_2+H_2}

\longrightarrow \sf{molMg=129gMg \times \dfrac{1molMg}{24.305gMg=} }

\small \sf{\longrightarrow molH_2=7.90molMg \times \dfrac{1molH_2}{1molMg}= 7.90molH_2 }

\small \sf{\longrightarrow massH_2=7.90H_2 \times \dfrac{2.016g}{1molH_2}= 7.90molH_2= 15.9 \: gH_2}

\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}

\small\bm{ The  \: mass \: of \:H_2 \: produce \: will \: be \: 15.9 } \bm{grams.}

mariarad [96]2 years ago
3 0

Answer:

Mg+2HCL-Magnesium Chloride +Hygrogen

>

Explanation:

N.O of moles=Mass\Molar Mass

192÷24=8

1:1

8×2=16

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marishachu [46]
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Each carbon atom covalently bonds with toms of carbon, hydrogen oxygen, and nitrogen

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What is the compound NH3
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Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.
Firlakuza [10]

Answer:

  • [Ag⁺] = 1.3 × 10⁻⁵M
  • s = 3.3 × 10⁻¹⁰ M
  • Because the common ion effect.

Explanation:

<u></u>

<u>1. Value of [Ag⁺]  in a saturated solution of AgCl in distilled water.</u>

The value of [Ag⁺]  in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

The ICE (initial, change, equilibrium) table is:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                    0

C          -s                      +s                  +s

E         X - s                   s                     s

Since s is very small, X - s is practically equal to X and is a constant, due to which the concentration of the solids do not appear in the Ksp equation.

Thus, the Ksp equation is:

  • Ksp = [Ag⁺] [Cl⁻]
  • Ksp = s × s
  • Ksp = s²

By substitution:

  • 1.8 × 10⁻¹⁰ = s²
  • s = 1.34 × 10⁻⁵M

Rounding to two significant figures:

  • [Ag⁺] = 1.3 × 10⁻⁵M ← answer

<u></u>

<u>2. Molar solubility of AgCl(s) in seawater</u>

Since, the conentration of Cl⁻ in seawater is 0.54 M you must introduce this as the initial concentration in the ECE table.

The new ICE table will be:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                  0.54

C          -s                      +s                  +s

E         X - s                   s                     s + 0.54

The new equation for the Ksp equation will be:

  • Ksp = [Ag⁺] [Cl⁻]
  • Ksp = s × ( s + 0.54)
  • Ksp = s² + 0.54s

By substitution:

  • 1.8 × 10⁻¹⁰ = s² + 0.54s
  • s² + 0.54s - 1.8 × 10⁻¹⁰ = 0

Now you must solve a quadratic equation.

Use the quadratic formula:

     

     s=\dfrac{-0.54\pm\sqrt{0.54^2-4(1)(-1.8\times 10^{-10})}}{2(1)}

The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer

<u>3. Why is AgCl(s) less soluble in seawater than in distilled water.</u>

AgCl(s) is less soluble in seawater than in distilled water because there are some Cl⁻ ions is seawater which shift the equilibrium to the left.

This is known as the common ion effect.

By LeChatelier's principle, you know that an increase in the concentrations of one of the substances that participate in the equilibrium displaces the reaction to the direction that minimizes this efect.

In the case of solubility reactions, this is known as the common ion effect: when the solution contains one of the ions that is formed by the solid reactant, the reaction will proceed in less proportion, i.e. less reactant can be dissolved.

3 0
3 years ago
A buret is filled with 0.1517 M A 25.0 mL portion of an unknown acid and two drops of indicator are added to an Erlenmeyer flask
erik [133]

Answer:

Molarity of Unknown Acid = 0.1332 M

Explanation:

Data for solving problem:

Molarity of base in buret (M₁)= 0.1517 M

volume of the acid in Erlenmeyer flask (V₂)= 25.0 mL

Volume of the base in the buret (V₁) = final volume of buret - initial volume in buret

final volume of buret = 22.5 mL

initial volume in buret = 0.55 mL

So

Volume of the base in the buret (V₁) = 22.5 mL -0.55 mL = 21.95 mL

Volume of the base in the buret (V₁)  = 21.95 mL

Molarity of Unknown acid in the Erlenmeyer flask (M₂) = To be find

Explanation:

It is acid base titration and  formula for this titration is as follows:

Molarity of base x Volume of base = Molarity of acid x volume of acid

it can be written as

M₁V₁ = M₂V₂ -------------------- equation (1)

we have to find M₂

so by rearrangment the equation (1)

M₁V₁ / V₂ = M₂ ------------------ equation (2)

put the values in equation in equation (2)

M₂ = 0.1517 M x 21.95 mL / 25.0 mL

M₂ = 3.3298 /25.0

M₂ = 0.1332 M

so the Molarity of Unknown acid is <u>0.1332 M</u>

5 0
3 years ago
Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,
WITCHER [35]

Answer: Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]

We are given:

\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ

Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

6 0
3 years ago
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