admin

Chemistry High School

If you start with 45.0 grams of ethylene and an excess of oxygen, how many grams of carbon dioxide will be produced?A.29 gB.57 gC.71 gD.141 g

1 minggu yang lalu

Solution 1

Guest #2270723

1 minggu yang lalu

Answer:

D.141 g

Explanation:

Given that:-

Mass of ethylene = 45.0 g

Molar mass of ethylene = 28.05 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{45.0\ g}{28.05\ g/mol}$

$Moles= 1.60\ mol$

According to the reaction below:-

$C_2H_4+3O_2\rightarrow 2CO_2+2H_2O$

1 mole of ethylene produces 2 moles of carbon dioxide

So,

1.60 mole of ethylene produces 2*1.60 moles of carbon dioxide

Moles of carbon dioxide = 3.2 mol

Molar mass of carbon dioxide = 44.01 g/mol

Mass = Moles*Molar mass = 3.2 mol x 44.01 g/mol = 141 g

D.141 g of carbon dioxide will be produced

📚 Related Questions

Question

What is the conecntration of Fe3 and the concentration of No3- present in the solution that result when 30.0 ml of 1.75M Fe(No3)3 are mixed with 45.0 ml of 1.00M Hcl?

Solution 1

Answer:

$[Fe^{+3}]=0.700 M$

$[NO_{3}^{-}]=2.10 M$

Explanation:

Here, a solution of Fe(NO₃)₃ is diluted, as the total volume of the solution has increased. The formula for dilution of the compound is mathematically expressed as:

$C_{1}. V_{1}= C_{2}.V_{2}$

Here, C and V are the concentration and volume respectively. The numbers at the subscript denote the initial and final values. The concentration of Fe(NO₃)₃ is 1.75 M. As ferric nitrate dissociates completely in water, the initial concentration of ferric is also 1.75 M.

Solving for [Fe],

$[Fe^{+3}]=\frac{C_{1}.V_{1}}{V_{2} }$

$[Fe^{+3}]=\frac{(1.75).(30.0)}{45.0+30.0 }$

$[Fe^{+3}]=0.700 M$

For [NO₃⁻],

There are three moles of nitrate is 1 mole of Fe(NO₃)₃. This means that the initial concentration of nitrate ions will be three times the concentration of ferric nitrate i.e., it will be 5.25 M.

$[NO_{3}^{-}]=\frac{C_{1}.V_{1}}{V_{2} }$

$[NO_{3}^{-}]=\frac{(5.25)(30.0)}{30.0+45.0 }$

$[NO_{3}^{-}]=2.10 M$

Question

A sample of argon gas has a volume of 500 mL at a pressure of 1.5 atm. If the pressure is increased to 4.0 atm, the new volume is given by

Solution 1

Answer:

187.5 mL

Explanation:

From the question we are given;

Initial volume of the gas, V₁ = 500 mL
Initial pressure of the gas, P₁ = 1.5 atm
New pressure of the gas, P₂ = 4.0 atm

We are required to determine the new volume of the gas;

We are going to use Boyle's law of gases.

According to Boyle's law, the volume of a fixed mass of a gas and pressure are inversely proportional at a constant absolute temperature.

That is;

$P\alpha\frac{1}{V}$

$P=\frac{k}{V}$

At different pressures and volume with constant temperature, then

$P_{1}V_{1}=P_{2}V_{2}$

Rearranging the formula;

$V_{2}=\frac{P_{1}V_{1}}{P_{2}}$

Thus;

$V_{2}=\frac{(500mL)(1.5 atm)}{4.0 atm}$

$=187.5 mL$

Thus, the new volume of the gas is 187.5 mL

Solution 2

Answer:

1.5x500

————-

4.0

Explanation:

Question

Methane gas is produced from the reaction of solid carbon and hydrogen gas: C(s)+2H2(g)→CH4(g) . How many liters of hydrogen gas at standard temperature and pressure (STP) are required to produce 40 liters of methane?

Solution 1

Answer:

80 liters

Explanation:

At STP, 1 mole of ideal gas has a volume of 22.4 liters.

Therefore, since liters and moles are directly proportional, we can use stoichiometry directly.

40L CH₄ × (2L H₂ / 1L CH₄) = 80L H₂

Question

For the most part, biological pH is slightly basic. However, the stomacłh is very acidic, and its contents must be swiftly neutralized by basic bicarbonate ions upon entering the small intestine, which has a basic pH. How many liters of 2 M Ba(OH)2 are needed to titrate a 4 Lsolution of 6 M H3PO4? a) 1.33 L b) 12L c) 18 L d) 56 L

Solution 1

Answer:

Vb = 18 L option c)

Explanation:

First, we need to write the titration reaction between the base and the acic, which is the following:

Ba(OH)₂ + H₃PO₄ <-------> Ba₃(PO₄)₂ + H₂O

However this equation is not balanced, we need to balance the equation adding some coefficients to the agents so:

3Ba(OH)₂ + 2H₃PO₄ <-------> Ba₃(PO₄)₂ + 6H₂O

Now that the equation is balanced, as we know this is an acid base titration, we need to calculate the mole ratio between the base and acid so:

moles B / moles A = 3/2

2 moles B = 3 moles A (1)

This is taken from the balanced reaction.

Now, finally we use the relation in titration which is:

moles A = moles B

or simply MaVa = MbVb

If we replace this in the ratio of this reaction we have:

2MbVb = 3MaVa (2)

And from there, we solve for Vb which is the volume of the base:

2 * 2 * Vb = 3 * 4 * 6

4Vb = 72

Vb = 72/4

Vb = 18 L

This is the volume of the base required to titrate this acid

Question

Fluorescent genes from a jellyfish can be inserted into bacteria with minor modifications, resulting in bacteria that can produce green fluorescent protein.

Solution 1

Answer:

Prokaryotes and eukaryotes use the same codons for translation.

Explanation:

Gene is the functional segment of the DNA and can be inherited. The particular gene codes for the particular protein that determines the morphology of the organisms.

The translation process is responsible for the expression of protein in prokaryotes and eukaryotes. The basic process of translation is quite similar. Genetic code is same in both prokaryotes and eukaryotes as the code is universal in nature and responsible for the production of protein in bacteria.

Solution 2

Consider this statement as True or false

Answer: True

Explanation:

Yes, it is possible that the gene responsible for the fluroscence in the jelly fish is extracted and inserted in the bacteria then the bacteria will also start glowing like jellyfish.

This is possible because the basic machinery is same in case of prokaryotes and eukaryotes.

The manipulation is known as genetic engineering by which the gene of interest from one organism is inserted in other organism.

Question

You have a total body concentration of 300 mOsM and total body volume of 3 liters. If you add 0.5 L of a solution containing 150 mosmol of NaCl to the body, what would the new total body concentration be?

Solution 1

Answer:

The new total body concentration would be 300 mOsM

Explanation:

In order to do this, we need to convert all concentrations to moles.

First, with the total body concentration, we have the initial volume of 3 liters and the concentration of 300 M (I will omit til the end the part of mOs)

The moles of the body concentration in this volume is:

moles = M * V

moles = 300 * 3 = 900 moles

To this moles, we add 150 moles of NaCL so, the total moles now is:

moles = 900 + 150 = 1050 moles

Finally, we can calculate the concentration with the new volume of 3.5 L (the sum of 3 and 0.5 liters added):

M = 1050 / 3.5

M = 300 mOsM

So the concentration remains the same as initial

Question

A sample of iron (Fe) with a mass of 200.0 g releases 9,840 cal when it freezes at its freezing point. What is the molar heat of fusion for iron?

Solution 1

The molar Heat of fusion based on the mass of iron and the heat released is 2756.3 cal/mol

What is the molar heat of the fusion of a substance?

The heat absorbed by one mole of a substance when it changes from a solid to a liquid is known as the molar heat of fusion (ΔHfus) of that substance. Any substance that melts absorbs heat, thus it seems sensible that a substance that freezes would release heat.

The molar heat of fusion is calculated as follows:

Heat change = moles * molar heat of fusion

moles of iron = mass / molar mass

molar mass of iron = 56 g/mol

mass of iron = 200 g

moles of iron = 200 / 56

moles of iron = 3.57 moles

Molar Heat of fusion = heat change / moles of iron

Molar Heat of fusion = 9840 / 3.57

Molar Heat of fusion = 2756.3 cal/mol

Learn more about molar heat of fusion at: brainly.com/question/29524112

#SPJ1

Question

How many fluoride ions (F-1) are needed to balance an aluminum ion (Al+3)?

Solution 1

WHen aluminum ions are added to fluoride ions, they form aluminum fluoride (AlF3). In this respect, when aluminum fluoride dissociates in water, it releases 3 fluoride ions and one aluminum ion. The answer is 3 fluoride ions.

Solution 2

Answer:

The correct answer is that we need three fluoride ions (F-1) to balance the compound.

Explanation:

Hello!

Let's solve this!

For the compound to be balanced, the charges have to be balanced.

We have +3 loads of Al and -1 of F.

We must multiply by 3 the Fluor so that it is as follows:

+ 3 + 3 * (- 1) = + 3-3 = 0

Therefore, we conclude that the correct answer is that we need three fluoride ions (F-1) to balance the compound.

Question

Which molecule,when added to a solution, would give the solution the highest ph? A. HCI B.H2SO4 C.KOH D.H2O

Solution 1

The correct answer among the choices given is option A. Hydrochloric acid is a much stronger acid than sulfuric acid given that the concentration both is equal. This can be proved by the pKa of these acids. Hydrochloric acid has a pKa of -6 while the pKa of sulfuric acid is -3. The lower the value of the pKa, the stronger the acidity of the substance.

Question

List the following aqueous solutions in order of decreasing freezing point: a. 0.040 m glycerin (C3H8O3) b. 0.020 m potassium bromide (KBr)c. 0.030 m phenol (C6H5OH).

Solution 1

Answer:c>b>a

Explanation:

Freezing point depends on relative molecular mass and nature of intermolecular forces. In glycerin, the relative molecular is the least of the the molecules are held together only by weak Vanderwaals forces. Phenol has stronger dipole interactions and higher relative molecular mass. Potassium bromide is ionic and posses very strong bonds hence it has the highest freezing point.

📂 Subjects

Mathematics 2647929

History 842281

English 748681

Biology 586256

Social Studies 406852

Chemistry 368373

Business 348603

Physics 324927

Health 199835

Spanish 130075

Geography 112100

Computers and Technology 106146

Arts 77164

Advanced Placement (AP) 23675

World Languages 23213

French 22589

Engineering 19607

Law 17108

Medicine 13966

SAT 10987

German 3389

🔢 Grades

High School 3423409

Middle School 2092250

College 1518097