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DaniilM [7]
4 years ago
6

The reaction represented by the equation: xA + yB ---> productsis started by mixing 1.00 mol of A with 2.00 mol of B in a 1.0

0 L container at a certain temperature. These concentrations are measured after a certain time.Substance Concentration, M[A] 0.875[B] 1.81Which is the x/y ratio in this reaction?
Chemistry
1 answer:
PIT_PIT [208]4 years ago
4 0

Answer:

The x/y ratio of the reaction is 0.68

Explanation:

Hi there!

The rate of dissapearence of A and B can be expressed as follows:

rate A = -Δ[A] /Δt

rate B = -Δ[B] / Δt

Where:

Δ[A] = change in the concentration of A (final [A] - initial [A])

Δ[B] = change in the concentration of B (final [B] - initial [B])

Δt = elapsed time.

These rates are equal if we divide each of them by their respective estechiometric coefficient:

1/x · (-Δ[A] /Δt) = 1/y · (-Δ[B] / Δt)

Let´s multiply by x each side of the equation:

-Δ[A] /Δt = x/y · (-Δ[B] / Δt)

now, let´s divide each side of the equation by (-Δ[B] / Δt)

(-Δ[A] /Δt) · (- Δt / Δ[B]) = x/y

Δ[A] / Δ[B] = x/y

Δ[A] = final [A] - initial [A] = 0.875 M - 1.00 M = -0.13 M

Δ[B] = final [B] - initial[B] = 1.81 M - 2.00 M = -0.19 M

-0.13 M / -0.19 M = x/y

x/y = 0.68

The x/y ratio of the reaction is 0.68

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son4ous [18]

Hey there!:

Molar mass CuSO4*5H2O = 249.68 g/mol

Therefore:

1 mole CuSO4*5H2O  ---------------- 249.68 g

3.2 moles --------------------------------- ?? ( mass of CuSO4*5H2O )

mass CuSO4*5H2O  = 249.68 * 3.2

mass CuSO4*5H2O = 798.97 g

Hope that helps!

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3 years ago
The chemical formula for a molecule that contains two chlorine (Cl) atoms is ___________.
Karo-lina-s [1.5K]
CI2 is two chlorine atoms.
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What is the name of the change when a liquid becomes a solid?​
Tcecarenko [31]

Answer:

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3 years ago
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Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M sodium hydroxide with 0.150 M HBr(aq). (
Taya2010 [7]

Answer:

a) pH = 13.176

b) pH = 13

c) pH = 12.574

d) pH = 7.0

e) pH = 1.46

f) pH = 1.21

Explanation:

HBr + NaOH ↔ NaBr + H2O

∴ equivalent point:

⇒ mol acid = mol base

⇒ (Va)*(0.150mol/L) = (0.025L)*(0.150mol/L)

⇒ Va = 0.025 L

a) before addition acid:

  • NaOH → Na+  + OH-

⇒ <em>C </em>NaOH = 0.150 M

⇒ [ OH- ] = 0.150 M

⇒ pOH = - Log ( 0.150 )

⇒ pOH = 0.824

⇒ pH = 14 - pOH

⇒ pH = 13.176

b) after addition 5mL HBr:

⇒ <em>C </em>NaOH = (( 0.025)*(0.150) - (0.005)*(0.150)) / (0.025 + 0.005) = 0.1 M

⇒ <em>C </em>HBr = (0.005)*(0.150) / ( 0.03 ) = 0.025 M

⇒ [ OH- ] = 0.1 M

⇒ pOH = 1

⇒ pH = 13

c) after addition 15mL HBr:

⇒ <em>C </em>NaOH = ((0.025)*(0.150) - (0.015)*(0.150 ))/(0.04) = 0.0375 M

⇒ <em>C </em>HBr = ((0.015)*(0.150))/(0.04) = 0.0563 M

⇒ [ OH- ] = 0.0375 M

⇒ pOH = 1.426

⇒ pH = 12.574

d) after addition 25mL HBr:

equivalent point:

⇒ [ OH- ] = [ H3O+ ]

⇒ Kw = 1 E-14 = [ H3O+ ] * [ OH- ] = [ H3O+ ]²

⇒ [ H3O+ ] = 1 E-7

⇒ pH = 7.0

d) after addition 40mL HBr:

⇒ <em>C</em> HBr = ((0.04)*(0.150) - (0.025)*(0.150)) / (0.04 + 0.025) = 0.035 M

⇒ [ H3O+ ] = 0.035 M

⇒ pH = 1.46

d) after addition 60mL HBr:

⇒ <em>C</em> HBr = ((0.06)*(0.150) - (0.025)*(0.150)) / (0.06+0.025) = 0.062 M

⇒ [ H3O+ ] = 0.062 M

⇒ pH = 1.21

8 0
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Answer:

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Explanation:

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8 0
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