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3 years ago

If 1.00 g of a hydrocarbon is combusted and found to produce 3.14 g of co2, what is the empirical formula of the hydrocarbon?

1 answer:

7 0

The combustion reaction is as expressed,

CxHy + O2 --> CO2 + H2O

The mass fraction of carbon in CO2 is 3/11. Hence,
mass of C in CO2 = (3.14 g)(3/11) = 0.86 g C.

Given that we have 1 g of the hydrocarbon, the mass of H is equal to 0.14 g.

moles of C = 0.86 g C / 12 g = 0.0713
moles of H = 0.14 g H / 1 g = 0.14

The empirical formula for the hydrocarbon is therefore, CH₂.

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Methanol is the simplest alcohol. How are larger alcohol molecules formed?

bazaltina [42]

Answer:

Methanol, also known as methyl alcohol amongst other names, is a chemical with the formula CH₃OH. It is a light, volatile, colourless, flammable liquid with a distinctive alcoholic odour similar to that of ethanol.

Explanation:

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3 years ago

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tatyana61 [14]

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4 years ago

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Which two of the following correctly describe a solution?

Mrac [35]

Answer:

C and A

Explanation:

i think it is because a solution is a mix of a solute and solvent, dont judge me if its wrong

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2 years ago

A 1.25g sample of dry ice is added to a 765mL flask containing nitrogen gas at a temperature of 25.0°C and a pressure of 725 mmH

erica [24]

Answer : The total pressure in the flask is 1.86 atm.

Explanation :

First we have to calculate the pressure of $CO_2$ gas.

Using ideal gas equation :

$PV=nRT\\\\P_{CO_2}=\frac{w}{M}\frac{RT}{V}$

where,

P = Pressure of $CO_2$ gas = ?

V = Volume of $CO_2$ gas = 765 mL = 0.765 L (1 L = 1000 mL)

n = number of moles

w = mass of $CO_2$ gas = 1.25 g

M = molar mass of $CO_2$ gas = 44 g/mol

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $CO_2$ gas = $25.0^oC=273+25.0=298K$

Putting values in above equation, we get:

$P_{CO_2}=\frac{w}{M}\frac{RT}{V}$

$P_{CO_2}=\frac{1.25g}{44g/mol}\frac{(0.0821L.atm/mol.K)\times 298K}{0.765L}=0.909atm$

Now we have to calculate the total pressure in the flask.

$P_T=P_{N_2}+P_{CO_2}$

Given :

$P_{CO_2}=0.909atm$

$P_{N_2}=725mmHg=\frac{725}{760}=0.954atm$

conversion used : (1 atm = 760 mmHg)

Now put all the given values in the above expression, we get:

$P_T=0.954atm+0.909atm=1.86atm$

Therefore, the total pressure in the flask is 1.86 atm.

5 0

3 years ago

For the Ag ground-state ion, predict the type of orbital from which an electron will need to be removed to form the ion of great

bagirrra123 [75]

Answer:

For the Ag ground-state ion, the type of orbital from which an electron will need to be removed to form the ion of greater positive charge is :

<u>s-orbital</u>

Explanation:

Naming of orbitals :

s = s-orbital

p = p-orbital

d = d-orbital

The electronic configuration of Ag is :

$[Kr]4d^{10}5s^{1}$

The Outermost shell = 5s

so electron should be removed from 5s . there is only 1 electron in 5s -shell. Hence , Ag exist +1 oxidation state.

According to Aufbau's Rule : Electron will be filled first in lower energy orbital.

The electron should be removed first ,from the higher energy orbital.This is because high energy orbitals are away from the nucleus and can be easily removed. The outermost electron feel less attraction of the nucleus

Ionization energy : Energy required to remove the electron from the outermost shell of the element in the gaseous state.

$Ag\rightarrow Ag^{+} +e^{-}$

Hence it will form Ag+ cation

$[Kr]4d^{10}$

Outer shell is removed after loss of electron.

4 0

4 years ago