Who thought everything was made from fire,earth,air and water

Answers

Answer 1

Answer:

Answer: Aristotles

Explanation:

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Work is the transfer of energy through motion. Force and distance are in the same directionA. True
B. False

Answers

Answer:

True

Explanation:

A sample of an unknown compound is vaporized at 190.°C. The gas produced has a volume of 2160.mL at a pressure of 1.00atm, and it weighs 2.87g. Assuming the gas behaves as an ideal gas under these conditions, calculate the molar mass of the compound. Be sure your answer has the correct number of significant digits.

Answers

Answer:

Molar mass = 50.4 g/mol

Explanation:

Pressure . Volume = number of moles (n) . R . T°K

This is the Ideal Gases Law where R is the Ideal Gases Constant and T°, Absolute Temperature.

We convert the T°C to T°K → 190°C + 273 = 463 K

R = 0.082 L.atm /mol.K

We need to convert the volume from mL to L → 2160 mL . 1L / 1000 mL = 2.16 L; now we replace:

1 atm . 2.16L = n . 0.082 L.atm /mol.K . 463K

(1 atm . 2.16L) / (0.082 L.atm /mol.K . 463K) = n → 0.569 moles

These moles refers to 2.87 g so let's find out the molar mass:

Molar mass (g/mol) = 2.87 g / 0.0569 mol = 50.4 g/mol

How many grams of oxygen are required to react with 10.0 grams of octane (C8H18) in the combustion of octane in gasoline?

Answers

Oh fun, stoichiometry. So the equation for this is C8H18+O2----> CO2+H2O, or
2C8H18+25O2----> 16CO2+18H2O when balanced. 10/114.229 g/mol = .087543... mol, converting to moles by dividing by molar mass. The ratio is 25/2, using the integers in front of the molecules in the balanced equation. so you just multiply .087543... by the ratio. That's 1.09429, which you multiply by the molar mass of O2, 31.9988 g/mol. The answer is 35.02 g O2

35.2 grams of oxygen are required to react with 10.0 grams of octane

$\boxed{\boxed{\bold{Further~explanation}}}$

Complete combustion of Hydrocarbons with Oxygen will be obtained by CO₂ and H₂O compounds.

If O₂ is insufficient there will be incomplete combustion produced by CO and H and O

Hydrocarbon combustion reactions (specifically alkanes)

$\large{\boxed {\bold {C_nH _ (_2_n _ + _ 2_) + \frac {3n + 1} {2} O_2 ----> nCO_2 + (n + 1) H_2O}}}$

For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:

Balancing C atoms, H and last atoms O atoms

Octane combustion reaction

C₈H₁₈ + O₂ ---> CO₂ + H₂O

To equalize the reaction equation we give the reaction coefficients with variables a, b and c while the most complex compounds, namely Octane, we give the number 1

So the reaction becomes

C₈H₁₈ + aO₂ ---> bCO₂ + cH₂O

C atom on the left 8, right b, so b = 8

left H atom = 18, right 2c so 2c = 18 ---> c = 9

Atom O on the left 2a, right 2b + 2c, so 2a = 2b + 2c

2a = 2.8 + 9

2a = 16 + 9

2a = 25

a = 25/2

The equation becomes:

C₈H₁₈ + 25/2O₂ ---> 8CO₂ + 9H₂O or be

2C₈H₁₈ + 25O₂ ---> 16CO₂ + 18H₂O

To find the mass O2, we find the mole first from the mole ratio with Octane

1. We are looking for the octane mole

Mr. octane = 8.Ar C + 18.Ar H

Mr. octane = 8.12 + 18.1

Mr octane = 96 + 18

Mr octane = 114

known octane mass: 10 grams then the mole:

mol = gram / Mr

mol = 10/ 114

mol = 0.088

2. We look for the O₂ mole

because of the ratio of the reaction coefficient between O2 and octane = 25: 2 then the mole of O₂ =

25/2 x 0.088 = 1,1

So that the mass O₂ =

mole. Mr = 1.1 32

mass = 35.2 grams

$\boxed{\boxed{\bold{Learn~more}}}$

a combustion reaction

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the type of chemical reaction

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The combustion of alkanes, alkenes, and alkynes

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Complete combustion of a sample of a hydrocarbon

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Keywords: Complete combustion of Hydrocarbons, alkanes, equalize the reaction equation

How many elements are in a set whole numbers between 3 and 15

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The set of elements composed of whole numbers between 3 and 15 consists of 4, 5, 6,7,8,9, 10, 11,12,13,14. Counting these, the number of elements is 11.

sodium chloride and glucose both are soluble in water but the solubility of NaCl is greater then glucose why.

Answers

The solubility of a substance in water depends on several factors, including the nature of the solute and solvent, as well as the intermolecular forces involved.

In the case of sodium chloride (NaCl) and glucose, both substances are indeed soluble in water. However, the solubility of NaCl is generally greater than that of glucose due to differences in their chemical properties.

NaCl is an ionic compound, meaning it consists of positive sodium ions (Na+) and negative chloride ions (Cl-). When NaCl is added to water, the polar water molecules surround the ions and separate them from the crystal lattice. These water molecules form favorable interactions with the charged ions through ion-dipole attractions, resulting in the dissolution of NaCl in water. These strong ion-dipole forces contribute to the high solubility of NaCl in water.

On the other hand, glucose is a covalent compound composed of carbon, hydrogen, and oxygen atoms. It does not dissociate into ions when dissolved in water. Instead, glucose molecules interact with water through weaker intermolecular forces such as hydrogen bonding and dipole-dipole interactions. While these forces allow glucose to dissolve in water, they are relatively weaker than the ion-dipole interactions in NaCl-water solution. As a result, glucose has a lower solubility compared to NaCl in water.

Therefore, the differing chemical properties of sodium chloride and glucose contribute to the higher solubility of NaCl in water compared to glucose.

I hope this helps.

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How much heat is required to vaporize 50.0 g of water if the initial temperature of the water is 25.0◦c and the water is heated to its boiling point where it is converted to steam? the specific heat capacity of water is 4.18 j · ( ◦c)−1 · g −1 and the standard enthalpy of vaporization of water at its boiling point is 40.7 kj · mol−1?

Answers

Q = mCΔT
Q is heat, m is mass, C is specific heat, and ΔT is change in temp

First, find how much energy you need to raise the heat to 100C, or the boiling point
Q = (50g)(4.18)(75 degrees) = 15675 J or 15.675 kJ

Then you need 40.7 kJ extra for each mole of water to evaporate it
If you have 50 grams of water and there are 18 grams per mol
50g / 18g = 2.8 moles x 40.7 = 113 kJ extra

15.675 + 113 kJ = 128.7 kJ required

Final answer:

The total heat required to heat 50 g of water from 25 °C to boiling and then vaporize it is approximately 128.821 kJ.

Explanation:

To calculate the total heat required to boil the water, we need to consider two steps: heating the water from its initial temperature to the boiling point, and then converting the water at its boiling point to steam.

Firstly, we calculate the heat needed to heat the water to the boiling point, using equation q = mcΔT: q1 = m * c * ΔT = 50.0 g * 4.18 J/(g • °C) * (100°C - 25°C) = 15675 J = 15.675 kJ
Next, amount of heat needed to vaporize the water at its boiling point can be found using the standard enthalpy of vaporization, ΔHvap = 40.7kJ/mol. But this value is per mole, so we need to convert grams of water to moles using the molar mass of water (18 g/mol). So, the number of moles of water is 50 g / 18 g/mol = 2.78 mol. The heat for vaporization, q2 = n * ΔHvap = 2.78 mol * 40.7 kJ/mol = 113.146 kJ.

Therefore, the total heat required to heat and vaporize the water is q = q1 + q2 = 15.675 kJ + 113.146 kJ = 128.821 kJ.

Learn more about Heat calculation here:

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