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blsea [12.9K]
2 years ago
6

A 0.271g sample of an unknown vapor occupies 294ml at 140C and 874mmHg. The emperical formula of the compound is CH2. How many m

oles of CO2 would be formed if 3moles of the compound react with oxygen?
Chemistry
1 answer:
Phoenix [80]2 years ago
7 0
Using PV = nRT, we can calculate the moles of the sample.
874 mmHg = 116,524 Pa
n = PV/RT
n = 116,524 x 294 x 10⁻⁶ / 8.314 x (140 + 273)
n = 9.98 x 10⁻³ mol

moles = mass / Mr
Mr = 0.271/9.98 x 10⁻³
Mr = 27.2
Mass of empirical formula = 14
Repeat units = 27.2 / 14 ≈ 2

Formula of substance:
C₂H₄

Combustion equation:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

1 mole produces 2 moles of CO₂, so 3 moles will produce 6 moles CO₂
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One of the most important chemical reactions is the Haber process, in which N2 and H2 are converted to ammonia which is used in
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Answer:

c) 22

Explanation:

Let's consider the following balanced equation.

N₂(g) + 3 H₂(g) ----> 2 NH₃(l)

According to the balanced equation, 34.0 g of NH₃ are produced by 1 mol of N₂. For 170 g of NH₃:

170gNH_{3}.\frac{1molN_{2}}{34.0gNH_{3}} =5.00molN_{2}

According to the balanced equation, 34.0 g of NH₃ are produced by 3 moles of H₂. For 170 g of NH₃:

170gNH_{3}.\frac{3molH_{2}}{34.0gNH_{3}} =15.0molH_{2}

The total gaseous moles before the reaction were 5.00 mol + 15.0 mol = 20.0 mol.

We can calculate the pressure (P) using the ideal gas equation.

P.V = n.R.T

where

V is the volume (50.0 L)

n is the number of moles (20.0 mol)

R is the ideal gas constant (0.08206atm.L/mol.K)

T is the absolute temperature (400.0 + 273.15 = 673.2K)

P=\frac{n.R.T}{V} =\frac{20.0mol\times (0.08206atm.L/mol.K)\times 673.2K ) }{50.0L} =22.0atm

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A farmer had an accident and spilled a chemical into his pond. Several days later, he notices many dead frogs around the pond. T
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the frogs limit the reproduction of the gnats, and therefore with less frogs the gnat population can increase
3 0
2 years ago
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The combustion of one mole of liquid ethanol, CH3CH2OH, produces 1367 kJ of heat. Calculate how much heat is produced when 235.0
deff fn [24]

Answer:- 6984 kJ of heat is produced.

Solution:- From given information, 1367 kJ of heat is produced by the combustion of 1 mole of ethanol. We are asked to calculate the heat produced by the combustion of 235.0 g of ethanol.

Let's convert given grams to moles and multiply by the heat produced by one mole of ethanol to get the total heat produced. Molar mass of ethanol is 46 grams per mole. The set will be:

235.0g(\frac{1mole}{46g})(\frac{1367 kJ}{1mole})

= 6984 kJ

So, 6984 kJ of heat is produced by the combustion of 235.0 g of liquid ethanol.

6 0
2 years ago
Sort these six events from the earliest to the most recent. Democritus proposes the existence of atoms- Dalton's atomic theory-
Igoryamba
These events are actually sorted right, according to the time they occurred.
1. Democritus proposes the existence of atoms - this happened in the 5th century BC
2. Dalton's atomic theory - it was first presented in 1803
3. J.J. Thomson discovers the electron - happened in 1897
4. Rutherford's gold foil experiment - somewhere between 1908 and 1913
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4 0
2 years ago
The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
QveST [7]

<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

7 0
2 years ago
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