While does kerosene float on water

liquid octane(CH3)(CH2)6CH3) reacts with gaseous oxygen gas(O2) to produce gaseous carbon dioxide(CO2) and gaseous water(H2O). I

max2010maxim [7]

The percent yield of carbon dioxide will be 49.0 %.

<h3>Percent yield</h3>

First, let's look at the equation of the reaction:

$2C_8H_1_8 + 25O_2 -- > 16CO_2 + 18H_2O$

The mole ratio of octane to oxygen is 2:25.

Mole of 3.43 g octane = 3.43/114.23 = 0.03 mol

Mole of 19.1 g oxygen = 19.1/32 = 0.60 mol

Thus, octane is limiting.

Mole ratio of octane to carbon dioxide = 2:16.

Equivalent mole of carbon dioxide = 0.03 x 8 = 0.24 mol

Mass of 0.24 mol carbon dioxide = 0.24 x 44.01 = 10.5624 grams

Percent yield of carbon dioxide = 5.18/10.5624 = 49.0 %

More on percent yield can be found here: brainly.com/question/17042787

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6 0

1 year ago

Two seventh grade students are studying how sound waves are transferring energy.

garik1379 [7]

Answer:

Hit the pot faster at a higher frequency

Explanation:

I feel like it would be because it makes more sense to me but I really have no clue tbh

6 0

3 years ago

Give the empirical formula for C↓8H↓8

mr_godi [17]

Answer:

Empirical formula of C₈H₈ = CH

Explanation:

Data Given:

Molecular Formula = C₈H₈

Empirical Formula = ?

Solution

Empirical Formula:

Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.

So,

tha ration of the molecular formula should be divided by whole number to get the simplest ratio of molecule

C₈H₈ Consist of Carbon (C), and Hydrogen (H)

Now

Look at the ratio of these two atoms in the compound

C : H

8 : 8

Divide the ratio by two to get simplest ratio

C : H

8/8 : 8/8

1 : 1

So for the empirical formula is the simplest ratio of carbon to hydrogen 1 : 1

So the empirical formula will be

Empirical formula of C₈H₈ = CH

4 0

3 years ago

How many mL of .450 M HCL is needed to neutralize 30mL of .150 M Ba(OH)2

Aleks04 [339]

You need .556M HCL to neutralize that

7 0

3 years ago

Part complete When a 235 92U nucleus is bombarded by neutrons (10n) it undergoes a fission reaction, resulting in the formation

nikdorinn [45]

Answer: The isotopic symbol of barium is $_{56}^{138}\textrm{Ba}$ and that of strontium is $_{38}^{89}\textrm{Sr}$

Explanation:

Nuclear fission reactions are defined as the reactions in which a heavier nuclei breaks down in two or more smaller nuclei.

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Ba}+^{94}_{36}\textrm{Kr}+3^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 3

A = 139

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 36 + 0

Z = 56

The isotopic symbol of barium is $_{56}^{139}\textrm{Ba}$

For the given fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Sr}+^{143}_{54}\textrm{Xe}+3^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 143 + 3

A = 90

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 54 + 0

Z = 38

The isotopic symbol of strontium is $_{38}^{89}\textrm{Sr}$

Hence, the isotopic symbol of barium is $_{56}^{138}\textrm{Ba}$ and that of strontium is $_{38}^{89}\textrm{Sr}$

7 0

3 years ago

While does kerosene float on water​

While does kerosene float on water