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2 years ago

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Consider the formation of nitryl fluoride: 2NO2(g)+F2(g)⇌2NO2F(g) The reaction is first order in F2 and second order overall. Wh

at is the rate law? View Available Hint(s) Consider the formation of nitryl fluoride: The reaction is first order in and second order overall. What is the rate law? rate=k[NO2]2[F2]2 rate=k[NO2][F2]2 rate=k[F2] rate=k[NO2] rate=k[NO2][F2] rate=k[NO2]2[F2]

1 answer:

Dahasolnce [82]2 years ago

7 0

Answer:

Rate = $k[NO_{2}][F_{2}]$

Explanation:

Two reactants are present in this reaction which are $NO_{2}and F_{2}$
We know overall order of a reaction is summation of individual order with respect to reactants present in rate law equation.
Here, overall order of reaction is 2 including first order with respect to $F_{2}$
So, rate of reaction should also be first order with respect to another reactant i.e. first order with respect to $NO_{2}$
So, rate law: rate = $k[NO_{2}][F_{2}]$

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Combustion of 8.652 g of a compound containing c, h, o, and n yields 11.088 g of coz, 3.780 g of h2o, and 3.864 g of no2. How ma

Katyanochek1 [597]

Answer:- C = 3.024 g, H = 0.42 g, N = 1.176 g and O = 4.032 g

Solution:- The compound contains C, H, N and O. On combustion, all the carbon is converted to carbon dioxide, All hydrogen is converted to water and all the nitrogen is converted to nitrogen dioxide.

From the grams of all these, we could calculate their moles and then using mol ratio of these products and the number of moles of C, H and N present in them, we calculate the moles of C, H and N respectively. Further, these moles are converted to grams. On subtracting the sum of grams of C, H and N from the mass of the sample, the mass of oxygen is calculated.

The calculations are as follows:

Calculations for the grams of C:-

$11.088gCO_2(\frac{1molCO_2}{44gCO_2})(\frac{1molC}{1molCO_2})(\frac{12gC}{1molC})$

= 3.024 g C

Calculations for the grams of H:-

$3.780gH_2O(\frac{1molH_2O}{18gH_2O})(\frac{2molH}{1molH_2O})(\frac{1gH}{1molH})$

= 0.42 g H

Calculations for the grams of N:-

$3.864gNO_2(\frac{1molNO_2}{46gNO_2})(\frac{1molN}{1molNO_2})(\frac{14gN}{1molN})$

= 1.176 g N

Mass of the compound is given as 8.652 g. Now we could calculate the grams of oxygen as:

mass of oxygen = 8.652 - (3.024 + 0.42 + 1.176)

= 8.652 - 4.62

= 4.032 g

So, 8.652 grams of the compound contains 3.024 g of C, 0.42 g of H, 1.176 g of N and 4.032 g of O.

7 0

2 years ago

Alka‑Seltzer is marketed as a remedy for stomach problems, such as heartburn or indigestion, and pain relief. It contains aspiri

Papessa [141]

Answer:

The equation for the reaction of one sodium bicarbonate ( NaHCO3 ) molecule with one citric acid (C6H8O7) molecule is the following:

Sodium Bicarbonate + Citric Acid ⇒ Water + Carbon Dioxide + Sodium Citrate

NaHCO3 + C6H8O7 ⇒ 3 CO2 + 3 H2O + Na3C6H5O7

Explanation:

The reaction is in balance, that is, the whole H2CO3 is not finished, but a little bit of this acid is left in the solution. Therefore, when sodium bicarbonate is added to the solution with citric acid, sodium citrate salt (C6H5O7Na3) and carbonic acid (H2CO3) are formed, which is rapidly broken down into water (H2O) and carbonic oxide (CO2).

C6H8O7 + NaHCO3 ⇒ C6H5O7Na3 + 3 H2CO3

C6H5O7Na3 + 3 H2CO3 ⇔ C6H5O7Na3 + 3 H2O + 3 CO2

5 0

2 years ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

Kisachek [45]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is $C_9H_{12}O$ and $C_{18}H_{24}O_2$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=39.01g$

Mass of $H_2O=10.65g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.01 g of carbon dioxide, $\frac{12}{44}\times 39.01=10.64g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 10.65 g of water, $\frac{2}{18}\times 10.65=1.18g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.64 + 1.18) = 1.6 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.64g}{12g/mole}=0.886moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.18g}{1g/mole}=1.18moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.6g}{16g/mole}=0.1moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.1 moles.

For Carbon = $\frac{0.886}{0.1}=8.86\approx 9$

For Hydrogen = $\frac{1.18}{0.1}=11.8\approx 12$

For Oxygen = $\frac{0.1}{0.1}=1.99\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 12 : 1

The empirical formula for the given compound is $C_9H_{12}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 272.38 g/mol

Mass of empirical formula = 136 g/mol

Putting values in above equation, we get:

$n=\frac{272.38g/mol}{136g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 9)}H_{(2\times 12)}O_{(2\times 1)}=C_{18}H_{24}O_2$

Hence, the empirical and molecular formula for the given organic compound is $C_9H_{12}O$ and $C_{18}H_{24}O_2$

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2 years ago

Bonds between two atoms that are equally electronegative are _____. bonds between two atoms that are equally electronegative are

Anna [14]

Bonds of two atoms of equal electronegativity are nonpolar covalent bonds.

Your second sentence is identical to the first sentence; I'll bet the second sentence is "Bonds between two atoms that are unequally electronegative are polar covalent bonds."

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2 years ago

Instrumental methods need only a microscopic sample to return an accurate result. Why is this so?

uranmaximum [27]

Instrumental methods of analysis rely on machines.The visualization of single molecules, single biological cells, biological tissues and nanomaterials is very important and attractive approach in analytical science.
There are several different types of instrumental analysis. Some are suitable for detecting and identifying elements, while others are better suited to compounds. In general, instrumental methods of analysis are:
-Fast
-Accurate (they reliably identify elements and compounds)
-Sensitive (they can detect very small amounts of a substance in a small amount of sample)

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2 years ago