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Ghella [55]
2 years ago
14

Write the balanced chemical equation between H2SO4H2SO4 and KOHKOH in aqueous solution. This is called a neutralization reaction

and will produce water and potassium sulfate. Phases are optional. balanced chemical equation: H2SO4+2KOH⟶K2SO4+2H2OH2SO4+2KOH⟶K2SO4+2H2O 0.750 L0.750 L of 0.470 M0.470 M H2SO4H2SO4 is mixed with 0.700 L0.700 L of 0.240 M0.240 M KOH.KOH. What concentration of sulfuric acid remains after eutralization?
Chemistry
1 answer:
const2013 [10]2 years ago
8 0

Answer:

0.185M sulfuric acid

Explanation:

Based on the reaction:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

<em>1 mole of sulfuric acid reacts with 2 moles of KOH</em>

Initial moles of H₂SO₄ and KOH are:

H₂SO₄: 0.750L ₓ (0.470mol / L) = <em>0.3525 moles of H₂SO₄</em>

KOH: 0.700L ₓ (0.240mol / L) = <em>0.168 moles of KOH</em>

The moles of sulfuric acis that react with KOH are:

0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.

Thus, moles that remain are:

0.3525moles - 0.0840 moles = <em>0.2685 moles of sulfuric acid remains</em>

As total volume is 0.700L + 0.750L = 1.450L, concentration is:

0.2685mol / 1.450L = <em>0.185M sulfuric acid</em>

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Vesnalui [34]

The concentrations : 0.15 M

pH=11.21

<h3>Further explanation</h3>

The ionization of ammonia in water :

NH₃+H₂O⇒NH₄OH

NH₃+H₂O⇒NH₄⁺ + OH⁻

The concentrations of all species present in the solution = 0.15 M

Kb=1.8 x 10⁻⁵

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\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}

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2 years ago
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Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine. CCl4(g) + (1/2)O
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The question is incomplete, here is the complete question:

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine.

CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c=4.4\times 10^9 at 1,000 K

Calculate Kc for the reaction 2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)

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<u>Explanation:</u>

The given chemical equations follows:

CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c

We need to calculate the equilibrium constant for the equation, which is:

2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)

As, the final reaction is the twice of the initial equation. So, the equilibrium constant for the final reaction will be the square of the initial equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c'=(K_c)^2

We are given:

K_c=4.4\times 10^9

Putting values in above equation, we get:

K_c'=(4.4\times 10^9)^2=1.936\times 10^{19}

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3 years ago
What are deltaTb and deltaTf for an aqueous solution that is 1.5g nacl in 0.250kg h2o? Given Kb=0.51 C/m and kr=1.86 C/m
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Answer:

T_f for given question is 2.79 and T_b is 0.52

\Delta T_b = I \times K_b \times m {i- vant hoff’s constant ; Kb- constant ; m molarity }

M = no. of moles of the solute present in one kg of solution

Let the weight of amount of solute be “w” and its molecular mass be “M”

Let the mass of the solvent in the given question be “x”

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\Delta T_b = I \times K_b \times w/Mx

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