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gayaneshka [121]
3 years ago
9

The reaction below can be catalyzed by manganese dioxide (MnO2). If you carried out such a reaction, which compound would you ex

pect to find in the reaction container at the end of the reaction?
2H2O2 ?2H2O + O2

A. MnO

B. MnH2

C. MnO2

D. MnOH
Chemistry
1 answer:
coldgirl [10]3 years ago
8 0

Answer:

\boxed{\text{C. MnO$_{2}$}}

Explanation:

2H₂O₂ ⟶ 2H₂O + O₂

A catalyst is a substance that speeds up a reaction but can be recovered unchanged at the end.

Thus, at the end of the reaction I would expect to find H₂O and MnO₂ in the container.

Water is not listed in any of the options, so I will have to choose \boxed{\text{C. MnO$_{2}$}}.

A, B, and D are wrong. They are not reactants, products, or catalysts.

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The energy of a photon is ________ proportional to its wavelength. The energy of a photon is ________ proportional to its freque
topjm [15]

Answer: Inversely  ,   Directly

Explanation:

The energy of a photon is inversely proportional to its wavelength and directly proportional to its frequency.

As can be seen from this equation;

E = hv  = h c / ∧

Where E = Energy of a photon

           v = Frequency

           h = Planck Constant

           c = speed of light

           ∧  = Wave length

4 0
3 years ago
How to balance this equation NH4OH+H3PO4=(NH4)3PO4+H2O with explainatio please ​
Maru [420]

Answer:

3 NH4OH (l) + H3PO4 (aq) → (NH4)3PO4 (aq) + 3 H2O (l)

Explanation:

This is an acid-base reaction (neutralization): NH4OH is a base, H3PO4 is an acid

5 0
2 years ago
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Balance the following equation.<br> ___ I + ___ O2 ___ I4O10
ANEK [815]
4I+5O2=I4O10? which would be a synthesis reaction
3 0
3 years ago
A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of t
Verdich [7]

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield  = 87.72%

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

  • moles of PbI₂
  • Limiting reactant
  • % yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

=  1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

7 0
3 years ago
Which statements regarding the Henderson-Hasselbalch equation are true? If the pH of the solution is known as is the pKa for the
Sonbull [250]

Answer:

1, 2, and 3 are true.

Explanation:

The Henderson-Hasselbalch equation is:

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

  • If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If you know pH and pka:

10^(pH-pka) = \frac{[A^-]}{[HA]}

The ratio will be: 10^(pH-pka)

  • At pH = pKa for an acid, [conjugate base] = [acid] in solution. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

0 = log₁₀ \frac{[A^-]}{[HA]}

10^0 = \frac{[A^-]}{[HA]}

1 = \frac{[A^-]}{[HA]}

As ratio is 1,  [conjugate base] = [acid] in solution.

  • At pH >> pKa for an acid, the acid will be mostly ionized. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH >> pKa,  10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]

  • At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH << pKa,  10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]

I hope it helps!

6 0
3 years ago
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