3 years ago

How many molecules are in a mole 0.500 mile of N205

Chemistry

1 answer:

kirill [66]3 years ago

5 0

D answer is D and that is your answer

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What is the maximum amount in moles of P2O5P2O5 that can theoretically be made from 112 gg of O2O2 and excess phosphorus

Nastasia [14]

Answer:

$n_{P_2O_5}^{max}=1.4molP_2O_5$

Explanation:

Hello!

In this case, since the reaction between phosphorous and oxygen to form diphosphorous pentoxide is:

$2P+\frac{5}{2}O_2\rightarrow P_2O_5$

Thus, since phosphorous is in excess and oxygen and diphosphorous pentoxide are in a 5/2:1 mole ratio, we can compute the maximum moles of product as shown below:

$n_{P_2O_5}^{max}=112 gO_2*\frac{1molO_2}{32.00gO_2}*\frac{1molP_2O_5}{5/2molO_2}\\\\ n_{P_2O_5}^{max}=1.4molP_2O_5$

Best regards!

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3 years ago

In your video respond to the following question/promt. Make sure you state a clear claim and explain with

Pavlova-9 [17]

Answer:

Vvvbn n njjnjj

Explanation:

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3 years ago

Read 2 more answers

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

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