Answer: He theorized that all material bodies are made up of indivisibly small “atoms.” Aristotle famously rejected atomism in On Generation and Corruption. Aristotle refused to believe that the whole of reality is reducible to a system of atoms.
Explanation:
Aristotle disagreed with Democritus primarily on the nature of matter and the concept of atomism.
Democritus proposed the theory of atomism, suggesting that all matter is composed of indivisible and indestructible particles called atoms. He believed that atoms combine and interact to form different substances.
On the other hand, Aristotle rejected the notion of atomism and instead proposed a different theory of matter. According to Aristotle, matter is continuous and infinitely divisible. He believed that substances are composed of four primary elements - earth, air, fire, and water - which have different inherent properties and undergo transformations through the combination and separation of these elements.
Aristotle's disagreement with Democritus stemmed from their differing views on the fundamental nature of matter. While Democritus emphasized the discrete and fundamental nature of atoms, Aristotle emphasized the continuous and composite nature of matter. Their contrasting viewpoints reflected a broader philosophical divide between atomism and holism in ancient Greek philosophy.
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Of the following, which is NOT a component of the circulatory system?
A. Heart
B. Veins
ws
C. Arteries
D. Brain
Answer:
D. Brain
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An ideal gas had a mass of 0.0218g and occupied a volume of 1.111L at 0oC and 0.0100atm. What is the molar mass of this gas?
Answer:
Explanation:
we shall use gas law formula to solve the problem .
P = .01 atm
V = 1.111 L
T = 273 K
mass of gas = .0218
molar mass = M
no of moles n = .0218 / M
PV = n RT
.01 x 1.111 = (.0218 / M) x .082 x 273
.0218 / M = .000496
M = .0218 / .000496
= 43.95
What does the word subcellular mean
Answer:
the word subcellular means contained within a cell. at a level of organization lower than the cellular.
Explanation:
2. If you travel 7.5 km and walk for 1.5 h. what is your average speed
Show Your Work Actual
Answer:
5km/h
Explanation:
Given parameters:
Distance = 7.5km
Time = 1.5h
Unknown:
Average speed = ?
Solution:
Average speed is the distance divided by the time taken.
Average speed = [tex]\frac{total distance }{time taken}[/tex]
Insert the parameters and solve;
Average speed = [tex]\frac{7.5}{1.5}[/tex] = 5km/h
How many different E2 products are expected in the reaction of 3-bromo-1,1-dimethylcyclohexane with NaOCH2CH3?
A) only 1
B) 2
C) 3
D) 4
Describe the three kinds of particles found in atoms. Where are they located in the atom and what are their charges?
Based on molecular orbital theory, the only molecule in the list below that has unpaired electrons is __________.a) C2b) N2c) F2d) O2e) Ne2
Answer:
B.
Explanation:
ive done this before trust
PLease help me I am giving a lot of points for this please help me
Answer:
Under balanced its motion under motion its s = d/t under s - d/t is time d = distance s= speed
Explanation:
Redox reactions stand for oxidation/reduction reactions. True or false: an oxidation reaction is always paired with a reduction reaction.
Answer:
The answer is: true
Explanation:
In redox reactions, the half-reactions of oxidation and reduction always occur simultaneously in pair.
The oxidation half-reaction involved the lost of electrons from a reduced substance (A) to form a oxidized substance (A⁺):
A ⇒ A⁺ + e-
In contrapossition, during the reduction half-reaction the oxidized substance (B⁺) gains electrons to form the reduced subtance (B):
B⁺ + e- ⇒ B
The overall redox reaction is obtained by the addition of the two half-reactions:
A ⇒ A⁺ + e-
B⁺ + e- ⇒ B
-----------------
A + B⁺⇒ A⁺ + B
The electrons gained by B are provided by A, which lost the same number of electrons. Thus, the oxidation/reduction reactions are paired.
Using the equations
2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)∆H° = -6271 kJ/mol
2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.6 kJ/mol
C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/mol
Determine the enthalpy (in kJ/mol) for the reaction
6 C (s) + 3 H₂ (g) → C₆H₆ (l).
The enthalpy : 49.1 kJ/mol
Further explanationThe change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
Based on the principle of Hess's Law, the change in enthalpy of a reaction will be the same even though it is through several stages or ways
Reaction
1. 2C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)∆H° = -6271 kJ/mol
Reverse
12 CO₂ (g) + 6 H₂O (g) ⇒ 2C₆H₆ (l) + 15 O₂ (g) ∆H° = 6271 kJ/mol : 2
6CO₂ (g) + 3H₂O (g) ⇒ C₆H₆ (l) + 15/2 O₂ (g) ∆H° = 3135.5 kJ/mol
2. 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.6 kJ/mol x 3/2
3H₂ (g) + 3/2O₂ (g) → 3H₂O (g) ∆H° = -725.4 kJ/mol
3. C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/mol x 6
6C (s) + 6O₂ (g) → 6CO₂ (g) ∆H° = -2361 k/j/mol
-------------------------------------------------------------------------------------
6 C (s) + 3 H₂ (g) → C₆H₆ (l) ∆H° = 49.1 kJ/mol
We add up and the same compound that is on different sides we eliminate
what is the molecular geometry of c atom in ch3nh2
Answer:
compound is sp3.
Explanation:
15.0 moles of gas are in a 8.00 L tank at 22.3 ∘C∘C . Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300L2⋅atm/mol2 a=2.300L2⋅atm/mol2 and b=0.0430L/molb=0.0430L/mol.
Answer:
[tex]\Delta P=4.10atm[/tex]
Explanation:
Hello!
In this case, since the ideal gas equation is used under the assumption of no interaction between molecules and perfectly sphere-shaped molecules but the van der Waals equation actually includes those effects, we can compute each pressure as shown below, considering the temperature in kelvins (22.3+273.15=295.45K):
[tex]P^{ideal}=\frac{nRT}{V}=\frac{15.0mol*0.08206\frac{atm*L}{mol*K}*295.45K}{8.00L}=45.5atm[/tex]
Next, since the VdW equation requires the molar volume, we proceed as shown below:
[tex]v=\frac{8.00L}{15.0mol}=0.533\frac{L}{mol}[/tex]
Now, we use its definition:
[tex]P^{VdW}=\frac{RT}{v-b} -\frac{a}{v^2}[/tex]
Thus, by plugging in we obtain:
[tex]P^{VdW}=\frac{0.08206\frac{atm*L}{mol*K}*295.45K}{0.533mol/L-0.0430L/mol} -\frac{2.300L^2*atm/mol^2}{(0.533L/mol)^2}\\\\P^{VdW}=49.44atm-8.09atm\\\\P^{VdW}=41.4atm[/tex]
Thus, the pressure difference is:
[tex]\Delta P=45.5atm-41.4atm\\\\\Delta P=4.10atm[/tex]
Best regards!
Dinitrogen tetraoxide, a colorless gas, exists in equilibrium with nitrogen dioxide, a reddish brown gas. One way to represent this equilibrium is:
The question is incomplete, the complete question is shown in the image attached
Answer:
F
T
Explanation:
From the equilibrium equation;
N2O4(g) ⇄NO2(g)
We must have to remember that;
1) At equilibrium, the concentration of the species on both sides of the reaction equation may not necessarily be the same but must be held at a constant value because the rate of forward reaction equals the rate of reverse reaction.
2) The rate of forward reaction must equal the rate of reverse reaction.
If the rate of forward reaction is k1 and the rate of reverse reaction is k2. Then it follows that at equilibrium k1 = k2.
Which two bonds are most similar in polarity? Which two bonds are most similar in polarity? O-F and Cl-F B-F and Cl-F I-Br and Si-Cl Al-Cl and I-Br C-Cl and Be-Cl
Answer:
A
Explanation:
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The branch of science which deals with chemicals and bonds is called chemistry.
The correct answer is A.
The polarity is defined as the separation of electric charge leading to a molecule or its chemical groups having an electric dipole moment, with a negatively charged end and a positively charged end.
The polar molecule has the following:-
Difference in electronegativity.In, the first compound the polarity is the same because the compound is the same that is O-F and CL-F.
Hence, the correct answer is A.
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Why does water have a much higher boiling point than methane even though water (H20) and methane (CH4) molecules are approximately the same size?
A. Only metallic bonds exist between methane molecules.
B. Only Van der Waals forces exist between water molecules.
C. Only metallic bonds exist between water molecules.
D. Only Van der Waals forces exist between methane molecules.
Answer:
Water has a higher boiling point because the hydrogen bonds that form water molecules are stronger than the Van der Waals interactions among methane molecules, therefor more energy must be provided in order to break the hydrogen bonds and allow the water molecules to escape the liquid state.
Explanation:
Water has a much higher boiling point than methane because only Van der Waals forces exist between methane molecules.
What are chemical bonds?Chemical bonds are forces of attraction between ions, molecules, and atoms.
There are different types of bonds present
Covalent bonds are the strongest bonds.
The molecules of water are joined by a hydrogen bond, thus its boiling point is high.
The molecules of Methane are attached by the van der Waals force, and it is weaker than the hydrogen bond.
Thus, methane has a lower boiling point than water.
Weaker bonds easily get broken.
Thus, the correct option is D.
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A student is preparing for the titration of 20.0 mL of an approximately 0.3 M solution of NH3 using HCl . She has a 50.0 mL buret and four possible HCl solutions to fill it with. Which solution would be best for use in her titration?a. 6.00 M HCI(aq) b. 0.3 M HCI(aq) c. 0.200 M HCI(aq) d. 0.100 M HCl(aq)
Answer:
c. 0.200 M HCI(aq)
Explanation:
The titration of a base as NH3 with hydrochloric acid is:
NH3 + HCl → NH4Cl
Where 1 mole of NH3 reacts per mole of HCl
The recommended volume to read in a burette is between 60 and 80% of its capacity.
That is 60% 50.0mL = 30mL and 80% 50.0mL = 40mL must be spent.
A 6.00M HCl spend:
20mL * (0.3mol/L) * (1L/ 6.00M) = 1mL. This solution is not the ideal for the titration.
A 0.3M HCl spend:
20mL * (0.3mol/L) * (1L/ 0.3M) = 20mL. This solution is not the ideal for the titration.
A 0.200M HCl spend:
20mL * (0.3mol/L) * (1L/ 0.200M) = 30mL. This solution ideal for the titration.
A 0.100M HCl spend:
20mL * (0.3mol/L) * (1L/ 0.100M) = 60mL. This solution is not the ideal for the titration.
Right option is:
c. 0.200 M HCI(aq)amu.
An atom with 4 protons, 5 neutrons, and 4 electrons has an atomic mass of
(Enter a whole number.)
Answer:
Explanation:
the sum of number of proton and neutron is known as atomic mass
so if number of proton is 4 and neutron is 5 then atomic mass of an atom is 9
2 or more forces that are not equal and opposite of one another can cause change in motion
A. Balanced force
B.friction
C.gravitational force
D.unbalanced force
Answer:
D unbalanced force
Explanation:
i think
What is the concentration of chloride ions in a solution that contains 0.375 g of aluminum chloride in 15.0 mL of solution?
Answer:
0.561 M
Explanation:
Step 1: Given data
Mass of aluminum chloride (m): 0.375 gMolar mass of aluminum chloride (M): 133.34 g/molVolume of the solution (V): 15.0 mLStep 2: Calculate the molar concentration of aluminum chloride (C)
We will use the following expression.
[tex]C = \frac{m}{M \times V(L) } = \frac{0.375g}{133.34 g/mol \times 0.0150L } = 0.187M[/tex]
Step 3: Write the reaction of dissociation of aluminum chloride
AlCl₃(aq) ⇒ Al³⁺(aq) + 3 Cl⁻(aq)
Step 4: Calculate the concentration of chloride ions
The molar ratio of AlCl₃ to Cl⁻ is 1:3. The concentration of Cl⁻ is 3/1 × 0.187 M = 0.561 M
how many moles of Na2SO4 are in 0.140 m aqueous solution that was prepared with 2.30 kg of water?
Answer:
0.322 mols of NaSO4
Explanation:
Molality = Mols of Solute / kg of Solvent
Molality is given as 0.140 m = mols/kg
kg of solution is given as 2.30 kg
Mols of Solute = Molality * kg of Solvent
Mols of Solute = [tex]\frac{0.140 }{ 1 kg } *2.30 kg[/tex]
Mols of Solute = 0.322 mols
The number of mole of Na₂SO₄ required to prepare the solution is 0.322 mole
The Molality of a solution is defined as the mole of solute per Kg of water i.e
Molality = mole / mass (Kg) of water.
With the above formula, we can obtain the mole of Na₂SO₄ in the solution as follow:
Molality of Na₂SO₄ = 0.140 M
Mass of water = 2.30 Kg
Mole of Na₂SO₄ =.?Mole = Molality × mass of water
Mole of Na₂SO₄ = 0.140 × 2.30
Mole of Na₂SO₄ = 0.322 moleTherefore, the mole of Na₂SO₄ in the solution is 0.322 mole
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During the formation of water from hydrogen atoms and one oxygen
Answer:
B
Explanation:
hydrogen only has one electron/molecule in it's only shell so it can get together and make a bond
Why can’t you ice skate on a lake when it is not frozen?
A sample of 6.44 g of napthalene (C10H8) is dissolved in 80.1 g of benzene (C6H6). Calculate the percent by mass of napthalene in this solution.
Answer:
7.44%
Explanation:
[tex]\frac{6.44}{6.44+80.1} *100=7.44[/tex]
Does MgO have a structure like that of NaCl or ZnS? If the density of MgO is 3.58g/cm3, estimate the radius (in centimeters) of the O2â anions and of the Mg2+ cations.
Answer:
The radius of O^2- is 1.5*10^-8 cm and the radius of
Mg ^2+ is 6.2*10^-9 cm
Explanation:
Face centered cubic : 1/8 atoms on each edges + 1/2 atoms on each face
= [tex]\frac{1}{8}* 8 + \frac{1}{2}*6[/tex] = 4 effective atoms
MgO have a structure like NaCl forms a lattice of FCC.
density of the lattice = [tex]\frac{ z.M}{a^3*Na}[/tex]
z : no of atoms
M: mass of the atoms
a: radius of the atom
Na: Avogadro's number
a^3(radius) = [tex]\frac{ 4*40.3}{3.58*6.022*10^23}[/tex]
a^3 = 7.477 * 10^-23 cm^3
a = 4.21 * 10^-8 cm
Now calculating the anionic(ra O^2-) and cationic (rc Mg^2+)
in Fcc a = 2[tex]\sqrt{2} ra[/tex]
ra = 1.5*10^-8 cm
a = 2ra + 2rc
rc = a/2 -ra
rc = 6.2*10^-9 cm
The radius of O^2- is 1.5*10^-8 cm and the radius of Mg ^2+ is 6.2*10^-9 cm
PLEASE HELP!!!!
what happens when the ocean reaches it’s saturation level of carbon dioxide?
Hi!
When the ocean reaches its saturation level of carbon dioxide, a couple of things will happen:
First, more carbon dioxide will remain in the atmosphere, and second, the ocean's pH will drop.
Carbonate and bicarbonate ions in seawater act as a buffer system which keeps the ocean's pH levels stable. When CO2 dissolves in seawater, it reacts with the ocean's buffer system in such a way that it produces two hydrogen ions, which lowers the pH. As more and more CO2 reacts with the ocean's buffer system and the system becomes saturated, less atmospheric carbon dioxide will cross over into the ocean. This excess CO2 will remain in the atmosphere and contribute to global climate change.
Hope this helped!
When the ocean reaches it's saturation level of carbon dioxide more amount of carbon dioxide remains in the atmosphere and pH of the ocean drops.
What is saturation level?It is defined as the level or extent up-to which a substance is soluble in a particular solvent. Due to global warming ,there is an excessive release of carbon dioxide in atmosphere. The excessive amount cannot be soluble in the ocean water which leads to ocean acidification.
As a result of ocean acidification the pH of water in the ocean drops thus there is less amount of oxygen present for aquatic organisms which is needed to survive. When carbon dioxide dissolves in water it yields carbonic acid which further reduces the pH of water and thus making it unfit for aquatic environment.
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You have a two-liter sample of each of the following gases, all at 25 °C and one atmosphere pressure. Which sample will weigh the most?A) Ammonia.B) Argon.C) Carbon dioxide.D) Nitrogen dioxide.E) Oxygen.
Answer: D) Nitrogen Dioxide
Explanation:
The characteristics of moles allow you to find that the correct answer for which sample weighs more is:
D) Nitrogen Dioxide
The molecular weight of the substance is the weight of one mole of a material, where one mole is Avogadro's number of particles.
In all chemical reactions the relationships are given as a function of moles.
Let's find the molecular weight of each of the substances, we take the atomic weights from the periodic table
Element PA (gr / mol)
H 1
C 12
N 14
O 16
Ar 40
A) Ammonia with NH₃ formula
weight of a mole of molecules
MP = 14 + 3
PM = 17 gr / mol
B) Argon symbol Ar
It is a noble gas and does not react
PM = 40 gr / mol
C) Carbon dioxide formula CO₂
Weight per mole of substance
PM = 12 + 2 16
Pm = 44 gr / mol
D) Nitrogen dioxide formula NO₂
Weight per molecular
MP = 14 + 2 16
PM = 46 gr / mol
E) Oxygen symbol O formula O₂
Molecular weights
PM = 2 16
PPM = 32 gr / mol
It indicates that we have two liters of sample, so the number of moles of each substance is the same, therefore the substance that weighs more is Nitrogen Dioxide (NO₂)
When reviewing the different answer, the correct one is:
D) Nitrogen Dioxide
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Blood takes about 1.55 s to pass through a 2.00 mm long capillary. If the diameter of the capillary is 5.00 μm and the pressure drop is 2.45 kPa, calculate the viscosity ???? of blood. Assume laminar flow.
Answer:
The viscosity [tex]\mathbf{\eta = 7.416 \times 10^{-4} \ N.s/m^2}[/tex]
Explanation:
From the given information:
Time t = 1.55 s
The radius of capillary = 5.00 μm /2
The pressure drop ΔP = 2.45 kPa
The length of the capillary = 2.00 mm
∴
The viscosity of the blood flow can be calculated by using the formula:
[tex]\eta = \dfrac{r^2 \Delta P }{8Lv}[/tex]
where;
v = L/t
Then;
[tex]\eta = \dfrac{r^2 \Delta P }{8L(\dfrac{L}{t})}[/tex]
[tex]\eta = \dfrac{(\dfrac{5 \times 10^{-6} \ m}{2})^2(2.45 \times 10^3 \ Pa) }{8(2.0 \times 10^{-3} \ m ) (\dfrac{2.0 \times 10^{-3} \ m }{1.55 \ s })}[/tex]
[tex]\eta = 7.416 \times 10^{-4} \ Pa.s[/tex]
To (N.s/m²)
[tex]\mathbf{\eta = 7.416 \times 10^{-4} \ N.s/m^2}[/tex]
What is 50% of 36?
O 9
O 16
O 18
O 34
Answer:
18
Explanation:
36 times 0.5
Phenol (C6H5OH) is often used as an antiseptic in mouthwashes and throat lozenges. If a mouthwash has a phenol concentration of 1.5 g per 100 mL of solution, what is the molarity of phenol?
Answer:
[tex]M=0.16M[/tex]
Explanation:
Hello!
In this case, since the molarity is defined as moles of solute divided by liters of solution, since we have phenol with a molar mass of 94.12 g/mol, we can first compute the moles in 1.5 g of phenol:
[tex]n=1.5g*\frac{1mol}{94.12 g}=0.016mol[/tex]
Next, since 1000 mL = 1 L, we notice that the volume of the solution is 0.100 L and therefore, the molarity of such solution turns out:
[tex]M=\frac{n}{V}=\frac{0.016mol}{0.100L}\\\\M=0.16M[/tex]
Best regards!
In the third century B.C., King Hieron of Syracuse asked the famous mathematician Archimedes to determine if his crown was made of pure gold. This was a puzzling problem for Archimedes-he knew how to measure the weight of the crown, but how could he measure the volume? Archimedes solved the problem when he got into his bath and noticed the water spilling over the sides of the tub. He realized that the volume of the displaced water must be equal to the volume of the object placed into the water. Archimedes was so excited by. his discovery that he jumped out of the bath and ran through the streets shouting "Eureka!" a. How can you tell if a crown is made of solid gold? b. Gold is one of the densest substances known, with a density of 19.3 g/mL. If the gold in the crown was mixed with a less-valuable metal like bronze or copper, how would that affect its density? c. What is the relationship between the object density, the liquid density, and the tendency of the object to float? d. From the experimental results obtained by you in the lab of the density of the liquid and the object, could the object float in that liquid? Explain.
Answer:
See explanation
Explanation:
a. I can conclusively tell if the crown was made of gold by measuring its density. First the mass of the crown is measured on a weighing balance. The crown is now put into a given volume of water and the volume of water displaced is accurately measured. The density of the crown is computed as mass/volume of fluid displaced. If the density of the crown is 19.3 g/mL, then it is made of solid gold.
b) When less valuable metals such as bronze or copper is mixed with gold in the crown, the density of the crown decreases and the crown becomes more brittle.
c) An object will float in a liquid when the density of the object is less than the density of the liquid. Hence the tendency of an object to float in a liquid depends on the density of the object and the density of the liquid.
d) Even though i do not know the results from your experiment but as regards the decision as to whether the object will float in the given liquid or not, reference must be made to the measured density of the object as well as the given density of the liquid. If the object is less dense (from values of density obtained from the experiment) than the liquid, then the object will float in the liquid and vice versa.