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## Complete combustion of 7.40 g of a hydrocarbon produced 23.8 g of co2 and 8.11 g of h2o. what is the empirical formula for the hydrocarbon?

You have to assume that the hydrocarbon contains only C, H and, possibly, O.

Then, all the C in CO2 and the H in H2O come from the hydrocarbon.

a) Number of moles of C in 23.8 g of CO2.

molar mass of CO2 = 12.0 g + 2 * 16.0 g = 44.0 g / mol

number of moles of C = mass in grams / molar mass = 23.8 g / 44.0 g/mol = 0.5409 mol

b) number of moles of H in 8.11 g of H2O

number of moles = mass in grams / molar mass = 8.11 grams / 18.0 g/mol = 0.4506 mol

c) mass of O in the hydrocarbon, m of O:

m of O = 7.40 g – mass of C – mass of H

mass of C = number of moles * atomic mass = 0.5409 mol * 12 g/mol = 6.49 g

m of H = 0.4506 mol * 1 g /mol = 0.45 g

=> grams of O = 7.40 g – 6.49 g – 0.45 g = 0.46 g of O.

d) number of moles of O = mass / atomic mass

=> number of moles of O = 0.46 g / 16 g / mol = 0.02875 mol O.

e) Empirical formula

C: 0.5409 / 0.02875 = 18.8 ≈ 19

H: 0.4506 / 0.02875 = 15.7 ≈ 16

O: 0.02875 / 0.02875 = 1

=> Empirical formula = C19 H16 O

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## Combustion analysis of an unknown compound provides the following data: 73.5 grams carbon (C), 4.20 grams hydrogen (H) and 72.3 grams chlorine (Cl). What is the percent composition of each element in this compound?

This problem is to apply Roult’s Law.

Roult’s Law states that the vapor pressure, p, of a solution of a non-volatile solute is equal to the vapor pressure of the pure solvent, Po solv, times the mole fraction of the solvent, Xsolv

p = Xsolv * Po sol

X solv = number of moles of solvent / number of moles of solution

The solvent is water and the solute (not volatile) is glycerin.

Number of moles = mass in grams / molar mass

mass of water = 132 ml * 1 g/ml = 132 g

molar mass of water = 18 g/mol

=> number of moles of water = 132 g / 18 g/mol = 7.33333 mol

mass of glycerin = 27.2 g

molar mass of glycerin:, C3H8O3: 3 * 12 g/mol + 8 * 1 g/mol + 3*16 g/mol = 92 g/mol

number of moles of glycerin = 27.2g / 92 g/mol = 0.29565

total number of moles = 7.33333 moles + 0.29565 moles = 7.62898 moles

=> X solv = 7.33333 / 7.62898 = 0.96125

=> p = 0.96125 * 31.8 torr ≈ 30.57 torr ≈ 30.6 torr.

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## How much heat is released when 75 g of octane is burned completely if the enthalpy of combustion is -5,500 kJ/mol C8H18

How much heat is released when 75 g of octane is burned completely if the enthalpy of combustion is -5,500 kJ/mol C8H18

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## Match the term to its description. 1. deforestation 2. combustion 3. circulation 4. weathering 5. calcium carbonate a. the primary component of mollusk shells and coral skeletons b. the logging of trees, which removes producers from the carbon cycle c. the burning of fossil fuels to make energy d. the process which brings up nutrients from the deep ocean e. the wearing down of rocks over time

The integumentary system covers the surface of the body is true about the integumentry system.

Explanation:

The integumentary system includes the skin, nails, glands, hairs, and nerves. Its main role is to act as a barrier to protect the body from the outside environment. It also retains body fluids, regulates body temperature, eliminates waste products, and protects against disease.

The integumentary system is the largest organ in the body that covers the surface of the body.

Thus, The integumentary system covers the surface of the body is the true statement about the integumentry system.

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## A device that initiates and maintains a controlled nuclear fission chain reaction to produce energy for electricity is a a. geothermal nuclear apparatus. b. nuclear combustion chamber. c. nuclear reactor. d. nuclear electric engine. e. electrical fusion device

Refer to the figure shown below.

Charge q₁ = 0.5 nC = 0.5×10⁻⁹ C
Charge q₂ = 8 nC = 8×10⁻⁹ C
d = 1.2 m, the distance between the two charges.

x is the distance between the two charges, measured from the charge q₁.

From Coulomb’s Law,
The electric field generated along x by q₁ is
E₁ = k(q₁/x²)
The electric field generated along x by q₂ is
E₂ = -k[q₂/(d-x)²]
where
k = 8.988×10⁹ (N-m²)/C² is the Coulomb constant/

When the electric field along x is zero, then
E₁ + E₂ = 0
k[q₁/x² – q₂/(d-x)²] = 0

That is,
0.5/x² = 8/(1.2 – x)²
8x² = 0.5(1.2 – x)²
16x² = 1.44 – 2.4x + x²
15x² + 2.4x – 1.44 = 0

x = (1/30)*[-2.4 +/- √(5.76 + 86.4)]
= 0.24 or -0.4 m

Reject the negative answer to obtain
x = 0.24 m
d-x = 0.96 m

The electric field is zero between the charges so that
(a) It is at 0.24 m from the 0.5 nC charge, and
(b) It is at 0.96 m from the 8 nC charge.

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## The following is an example of which type of chemical reaction? AgNO3 (aq) + NaCl (aq) àAgCl (s) + NaNO3 (aq) single replacement synthesis double replacement combustion

Explanation:

It is known that general charge on a hydrogen atom is +1, oxygen atom is -2, and sodium atom is +1.

Hence, oxidation number of nitrogen atom in the given species or compounds is as follows.

• x + 4(1) = 1

x = 1 – 4

= -3

• 2x + 5(-2) = 0         (since total charge on the molecule is zero)

x = +5

• 1 + x + 3(-2) = 0

x = +5

Thus, we can conclude that the oxidation numbers of nitrogen in NH4+, N2O5, and NaNO3 are -3, +5, +5 respectively.

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## What type of reaction does the following equation represent? AgNO3 + NaCI > AgCI + NaNO3 A. Synthesis B. Combustion C. Single-displacement reaction D. Double-displacement reaction

Answer : The excess reagent is, Explanation : Given,

Mass of NaCl = 4 g

Mass of = 10 g

Molar mass of NaCl = 58.44 g/mole

Molar mass of = 169.87 g/mole

First we have to calculate the moles of and .  Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is, From the balanced reaction we conclude that

As, 1 moles of react with 1 mole of So, 0.059 moles of react with 0.059 mole of From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Hence the excess reagent is, Categories

## Name the compound that is a reactant in all combustion reactions.

Name the compound that is a reactant in all combustion reactions.