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timurjin [86]
3 years ago
10

Iron has a density of 7.86 g/cm3. Calculate the volume (in dL) of a piece of iron having a mass of 4.79 kg .

Chemistry
1 answer:
Kazeer [188]3 years ago
5 0
M = 4.79 kg = 4790 g
d = 7.86 g/cm3
density = mass/volume
v = m/d
v = 4790/7.86
v = 609.4 cm3
1 cm3 = 10^-2 dL

v = 609.4 x 0.01 = 6.09 dL
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Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
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damaskus [11]

The amount, in mg, of CO present in the room will be 191,520 mg.

<h3>Stoichiometric problem</h3>

The concentration of the gas in the room is 5.7 x 10^{-3} mg/cm3.

The dimension of the room is 3.5 m x 3.0 m x 3.2 m. This is equivalent to 350 cm x 300 cm x 320 cm.

We can obtain the volume of the room as:

                  350 x 300 x 320 = 33,600,000 cm3

The concentration is in mg/cm3, meaning that it is mass/volume.

Thus:

 mass = concentration x volume = 5.7 x 10^{-3} mg/cm3 x 33,600,000 cm3

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The mass of CO in the room is 191,520 mg

More on stoichiometric problems can be found here: brainly.com/question/14465605

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Alexeev081 [22]

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