Intermolecular forces dipole differences london dispersion

A sample of a compound that contains only the elements C, H, and N is completely burned in O₂ to produce 44.0 g of CO₂, 45.0 g o

koban [17]

Answer:

CH₅N

Explanation:

In the combustion, all of the C in the compound was used to produce CO₂ in a 1:1 ratio. Thus, the moles of CO₂ (MW 44.01 g/mol) produced equals the moles of C in the compound:

(44.0 g)(mol/44.01g) = 0.99977 mol CO₂ = 0.99977... mol C

Similarly, all of the H in the compound was used to produce H₂O in a ratio of 2H:1H₂O. The moles of H₂O (MW 18.02 g/mol) produced was:

(45.0 g)(mol/18.02g) = 2.497...mol H₂O

Moles of H is found using the molar ratio of 2H:1H₂O:

(2.497...mol H₂O)(2H/1H₂O) = 4.994...mol H

The ratio of H to C in the compound is:

(4.994...mol H)/(0.99977... mol C) = 5 H:C

Some NO₂ was produced from the N in the compound. Assuming a 1:1 ratio of C:N, the simplest empirical formula is: CH₅N.

3 0

2 years ago

An organic compound is 61.5% C, 2.56% H and 35.9% N by mass. 2.00 grams of this gas is entered into a 300.0 mL flask and heated

pashok25 [27]

Answer:

C₄H₂N₂

Explanation:

First we calculate the moles of the gas, using PV=nRT:

P = 2670 torr ⇒ 2670/760 = 3.51 atm

V = 300 mL ⇒ 300/1000 = 0.3 L

T = 228 °C ⇒ 228 + 273.16 = 501.16 K

3.51 atm * 0.3 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 501.16 K

n = 0.0256 mol

Now we calculate the molar mass of the compound:

2.00 g / 0.0256 mol = 78 g/mol

Finally we use the percentages given to calculate the empirical formula:

C ⇒ 78 g/mol * 61.5/100 ÷ 12g/mol = 4

H ⇒ 78 g/mol * 2.56/100 ÷ 1g/mol = 2

N ⇒ 78 g/mol * 35.9/100 ÷ 14g/mol = 2

So the empirical formula is C₄H₂N₂

6 0

2 years ago

In a first-order decomposition reaction, 50.0% of a compound decomposes in 10.5 min.

mihalych1998 [28]

In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol $t_{1/2}$ .

(a) For first order reaction, rate constant and half life time are related to each other as follows:

$k=\frac{0.6932}{t_{1/2}}=\frac{0.6932}{10.5 min}=0.066 min^{-1}$

Thus, rate constant of the reaction is $0.066 min^{-1}$ .

(b) Rate equation for first order reaction is as follows:

$k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}$

now, 75% of the compound is decomposed, if initial concentration $[A_{0} ]$ is 100 then concentration at time t $[A_{t} ]$ will be 100-75=25.

Putting the values,

$0.066 min^{-1}=\frac{2.303}{t}log\frac{100}{25}=\frac{2.303}{t}(0.6020)$

On rearranging,

$t=\frac{2.303\times 0.6020}{0.066 min^{-1}}=21 min$

Thus, time required for 75% decomposition is 21 min.

8 0

3 years ago

Sixty liters of a gas were collected over water when the barometer read 663 mmhg , and the temperature was 20∘c. what volume wou

lesya [120]

First, let's compute the number of moles in the system assuming ideal gas behavior.

PV = nRT
(663 mmHg)(1atm/760 mmHg)(60 L) = n(0.0821 L-atm/mol-K)(20+273 K)
Solving for n,
n = 2.176 moles

At standard conditions, the standard molar volume is 22.4 L/mol. Thus,

Standard volume = 22.4 L/mol * 2.176 mol = 48.74 L

3 0

3 years ago

What promotes greater hardness in minerals?

Mashutka [201]

Smaller atoms and stronger bonds promotes greater hardness in minerals.

7 0

2 years ago

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