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3 years ago

How many milliliters of 0.021 moles of oxalic acid is needed to react with 80 mL of 0.11 moles of KMnO4

Chemistry

1 answer:

alexira [117]3 years ago

8 0

Answer:

6.11 mL of H₂C₂O₄ are needed

Explanation:

We determine the reaction which is a redox one, in acidic medium

C₂O₄⁻² + MnO₄⁻ → CO₂ + Mn²⁺

C₂O₄⁻² → 2CO₂ + 2e⁻ Oxidation

Carbon changes the oxidation state, from +3 to +4

5e⁻ + MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O Reduction

We add 4 water to the product side, in order to balance the oxygen and, we have 8H+ in the reactant side, in order to balance the H

Mn changes the oxidation state from +7 to +2

(C₂O₄⁻² → 2CO₂ + 2e⁻) .5

5C₂O₄⁻² → 10CO₂ + 10e⁻

(5e⁻ + MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O) .2

10e⁻ + 2MnO₄⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O

10e⁻ + 2MnO₄⁻ + 16H⁺ + 5C₂O₄⁻² → 2Mn²⁺ + 8H₂O + 10CO₂ + 10e⁻

The electrons are cancelled, so the balanced reaction is:

2KMnO₄ + 6HCl + 5H₂C₂O₄ → 2MnCl₂ + 8H₂O + 10CO₂ + 2KCl

Concentration of KMnO₄ = 0.11 mol / 0.080mL = 1.375M

Imagine that the reactants are in molar concentration (mol/L)

Ratio in stoichiometry is 2:5

2 moles of KMnO₄ react to 5 moles of oxalic acid

Then, 1.375 moles of KMnO₄ will react to (1.375 moles . 5 )/ 2 = 3.437 M

Concentration of H₂C₂O₄ = 3.437 M (mol/L)

3.437 mol/L = 0.021 moles / Volume (L)

0.021 moles / 3.437 mol/L = Volume (L) → 0.00611 L

0.00611 L . 1000 mL / 1L = 6.11 mL

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3 years ago

Consider an electron with a mass of 9.11 x 1051 kg and a 100.0 g tennis ball that are both moving with a velocity of 70.0 m s1.

MrRissso [65]

Answer:

(a) 6.38 × 10⁻²⁹ kg·m·s⁻¹; (b) 7.00 kg·m·s⁻¹; (c) 82.7 µm; (d) 7.53 × 10⁻³⁴ m;

(e) Δx ∝ 1/m

Explanation:

(a) Momentum of electron

p = mv = 9.11 × 10⁻³¹ kg × 70.0 m·s⁻¹ = 6.38 × 10⁻²⁹ kg·m·s⁻¹

(b) Momentum of tennis ball

p = mv = 0.1000 kg × 70.0 m·s⁻¹ = 7.00 kg·m·s⁻¹

(c) Δx for electron

Δp = 0.010p = 0.010 × 6.38 × 10⁻²⁹ kg·m·s⁻¹ = 6.38 × 10⁻³¹ kg·m·s⁻¹

$\begin{array}{rcl}\Delta x \Delta p & \geq & \dfrac{h}{4 \pi}\\\\\Delta x \times 6.38 \times 10^{-31} \text{ kg$\cdot$m$\cdot$s$^{-1}$} & \geq & \dfrac{6.626 \times 10^{-34} \text{ kg$\cdot$m$^{2}$s}^{-1}}{4 \pi}\\\\\Delta x \times 6.38 \times 10^{-31} & \geq & 5.273 \times 10^{-35} \text{ m}\\\Delta x & \geq & \dfrac{5.273 \times 10^{-35} \text{ m}}{6.38 \times 10^{-31}}\\\\ & \geq&8.27 \times10^{-5} \text{ m}\\ &\geq&\textbf{82.7 $\mu$m}\\\end{array}$

(d) Δx for tennis ball

Δp = 0.010p = 0.010 × 7.00 kg·m·s⁻¹ = 0.0700 kg·m·s⁻¹

$\begin{array}{rcl}\Delta x \times 0.0700 & \geq & 5.273 \times 10^{-35} \text{ m}\\\Delta x & \geq & \dfrac{5.273 \times 10^{-35} \text{ m}}{0.0700}\\\\ &\geq& \textbf{7.53 $\mathbf{\times 10^{-34}}$ m}\\\end{array}$

(e) Relative uncertainty

Both particles are travelling at the same speed, so,

ΔxΔp = Δx × mv = mvΔx = constant

v is constant, so

Δx ∝ 1/m

Thus, the larger the mass of an object, the smaller the uncertainty in its velocity.

5 0

3 years ago

PLEASE PLEASE HELP 1. Using a neutral atom that has an atomic number of 28 and a mass number of 152, how many protons, electrons

vampirchik [111]

Answer:

The answer to your question is:

Explanation:

Data

Atomic number = Z = 28

Mass number = A = 152

Atomic number number = # of protons

Mass number = # of protons + # of neutrons

Then

Protons = 28

Electrons = 28

Neutrons = 152 - 28 = 124

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3 years ago

Can someone help me with a background research paper on what makes ice melt fastest, I am in need of help

MArishka [77]

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Effect of Sand on Ice
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Effect of Sugar on Ice
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While sugar will melt ice, it still isn’t as effective as salt, which breaks down into sodium and chloride ions. When one molecule of salt dissolves, it adds two components to the solution, providing more interference in preventing water molecules from freezing.

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notka56 [123]

Answer:

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