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2 years ago

Balance this equationPB3 O4 + H2 → PB + H2O

Chemistry

1 answer:

Anna35 [415]2 years ago

4 0

Answer:

Pb3O4 + 4H2 → 3Pb + 4H2O

Explanation:

Pb3O4

Tritium - H2

Molar Mass of H2 Bond Polarity H-3 Hydrogen-3 3H T

Products

Lead - Pb

Molar Mass of Pb Plumbum Element 82 Bulk Lead

Water - H2O

Molar Mass of H2O Oxidation Numbers of H2O Dihydrogen Monoxide Dihydridooxygen Hoh Hydrogen Hydroxide Dihydrogen Oxide Oxidane Hydrogen Oxide Pure Water

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The combustion of ethene in the presence of excess oxygen yields carbon dioxide and water: c2h4 (g) + 3o2 (g) → 2co2 (g) + 2h2o

meriva

Answer:

$\boxed{-267.5}$

Explanation:

You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.

The formula is

$\Delta_{r} S^{\circ} = \sum_n {nS_{\text{products}}^{\circ} - \sum_{m} {mS_{\text{reactants}}^{\circ}}}$

The equation for the reaction is

C₂H₄(g) + 3O₂(g) ⟶ 2CO₂(g) + 2H₂O(ℓ)

ΔS°/J·K⁻¹mol⁻¹ 219.5 205.0 213.6 69.9

$\Delta_{r} S^{\circ} = (2\times213.6 + 2\times69.9) - (1\times219.5 + 3\times205.0)\\\\= 567.0 - 834.5 = \boxed{-267.5 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1}}$

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3 years ago

How much heat (in kj) is released when 3.600 mol naoh(s) is dissolved in water? (the molar heat of solution of naoh is â445.1 kj

Fantom [35]

<span>-1602 kj

Sauce:me

Your welcome
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8 0

2 years ago

Read 2 more answers

55 points take the or whatever but Im getting head on my 15th bday on march 26th gonna be great also help me Two or more substan

Kipish [7]

Mixture is the answer.

3 0

2 years ago

Consider the following reaction:

adell [148]

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

Choose whether each statement is a scientific hypothesis, theory, or law. The pressure a gas exerts on a container increases as

Naya [18.7K]

Answer:

1) theory

2) law

3) hypothesis

Explanation:

edge 2021

7 0

2 years ago

Balance this equationPB3 O4 + H2 → PB + H2O​

Balance this equationPB3 O4 + H2 → PB + H2O