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s344n2d4d5 [400]
3 years ago
13

Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction: 2NO2 → 2NO + O2 In a particular experiment at 300 °C, [

NO2] drops from 0.0100 to 0.00650 M in 100 s. The rate of disappearance of NO2 for this period is __________.
Chemistry
1 answer:
Setler [38]3 years ago
3 0

Answer:

rate=-1.75x10^{-5}\frac{M}{s}

Explanation:

Hello,

In this case, for the given information, we can compute the rate of disappearance of NO₂ by using the following rate relationship:

rate=\frac{1}{2}*\frac{C_f-C_0}{t_f-t_0}

Whereas it is multiplied by the the inverse of the stoichiometric coefficient of NO₂ in the reaction that is 2. Moreover, the subscript <em>f</em> is referred to the final condition and the subscript <em>0</em> to the initial condition, thus, we obtain:

rate=\frac{1}{2}*\frac{0.00650M-0.0100M}{100s-0s}\\\\rate=-1.75x10^{-5}\frac{M}{s}

Clearly, it turns out negative since the concentration is diminishing due to its consumption.

Regards.

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What's the pH of a solution with a concentration of hydronium ions of 5.29x10^-10?
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A scientist measures the standard enthalpy change for the following reaction to be -327.2 kJ : P4O10(s) 6 H2O(l)4H3PO4(aq) Based
OleMash [197]

Answer:

ΔH°f P4O10(s) = - 3115.795 KJ/mol

Explanation:

  • P4O10(s) + 6H2O(l) ↔ 4H3PO4(aq)
  • ΔH°rxn = ∑νiΔH°fi

∴ ΔH°rxn = - 327.2 KJ

∴ ΔH°f H2O(l) = - 285.84 KJ/mol

∴ ΔH°F H3PO4(aq) = - 1289.5088 KJ/mol

⇒ ΔH°rxn = (4)(- 1289.5088) - (6)(- 285.84) - ΔH°f P4O10(s) = - 327.2 KJ

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5 0
3 years ago
A compound with the empirical formula CH2 was found to have a molar mass of approximately 112 g. Write the molecular formula of
san4es73 [151]

Answer:

C8H10

Explanation:

n (CH2) = 112

n (12 + 1 + 1) = 112

n (14) = 112

n = 8

Molecular Formula: C8H10

6 0
2 years ago
what is an empirical formulaWhat is the percent composition by mass of nitrogen in (NH4)2CO3 (gram-formula mass = 96.0 g/mol)?
Margarita [4]

Answer:

Percentage composition = 14.583%

Explanation:

In chemistry, the emprical formular of a compound is the simplest formular a compound can have. It shows the simplest ratio in which the elements are combined in the compound.

Percentage composition by mass of Nitrogen

Nitrogen = 14g/mol

In one mole of the compound;

Mass of Nitrogen = 1 mol * 14g/mol = 14g

Mass of compound = 1 mol * 96.0 g/mol = 96

Percentage composition of Nitrogen = (Mass of Nitrogen /  Mass of compound) * 100

percentage composition = 14/96   * 100

Percentage composition = 0.14583 * 100

Percentage composition = 14.583%

3 0
3 years ago
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