The volume of a sample of oxygen is 300mL when the pressure is 1 atm and the temperature is 27 C . At what temperature is the volume 1.00 L and the pressure.500 atm?

Answer:

question is not clear please send clear question

Answer:

linear

Explanation:

nothing bends off

Answer:

H3C—CH3

Explanation:

The strength of a bond is indicated by the value of its bond dissociation energy. Simply put, the bond dissociation energy is the energy required to break the bond.

Carbon forms single, double and triple bonds with itself. As a matter of fact, carbon atoms can link to each other indefinitely. This is known as catenation and has been attributed to the low bond energy of the carbon-carbon single bond.

The bond energy of the carbon-carbon single bond is about 90KJmol-1 while that of carbon-carbon double bond is about 174KJmok-1. The carbon-carbon triple bond has the highest bond dissociation energy of about 230KJmol-1.

Hence, it is easier to break carbon-carbon single bonds than double and triple bonds respectively, hence the answer.

According to the forces of attraction, the molecule which can be easily broken is CH₃-CH₃ as it has a single bond with low dissociation energy as compared to double or triple bonds.

Forces of attraction is a force by which atoms in a molecule  combine. it is basically an attractive force in nature.  It can act between an ion  and an atom as well.It varies for different  states  of matter that is solids, liquids and gases.

The forces of attraction are maximum in solids as  the molecules present in solid are tightly held while it is minimum in gases  as the molecules are far apart . The forces of attraction in liquids is intermediate of solids and gases.

The physical properties such as melting point, boiling point, density  are all dependent on forces of attraction which exists in the substances.Single bonds have least dissociation energy while triple bonds have the maximum  dissociation energy.

Thus,the molecule which can be easily broken is CH₃-CH₃.

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Answer:

Crystalline solids have well-defined edges and faces, diffract x-rays, and tend to have sharp melting points.

 In contrast, amorphous solids have irregular or curved surfaces, do not give well-resolved x-ray diffraction patterns, and melt over a wide range of temperatures.

Answer:

a) the percentage yield will exceed 100%

b) the excess reactant is filtered along with the barium sulphate precipitate. It is possible to recover the excess reactant by carefully washing the precipitate with water.

Explanation:

In the precipitation of barium sulphate, the ions in the reactants exchange partners in the product leading to an insoluble product.

In every reaction, there is a limiting reactant whose amount determines the amount of product that can be obtained. The reactant in excess remains in the system even after the reaction is completed and may be recovered alongside the product which leads to a percentage yield above 100%.

If the excess reactant is soluble in water, it can be recovered from the precipitate if needed by washing the precipitate with water.

Answer:

By a factor of 12

Explanation:

For the reaction;

A + 2B → products

The rate law is;

rate = k[A]²[B]

As you can see, the rate is proportional to the square of the concentration of  A  and the of the concentration of  B .

Let's say initially, [A] = x, [B] = y

The rate law in this case is equal to;

rate1 = k. x².y

Now you double the concentration of A and triple the concentration of B.

[A] = 2x, [B] = 3y

The new rate law is given as;

rate2 = k . (2x)². (3y)

rate2 = k . 4x² . 3y

rate2 = 12 k . x² . y

Comparing rate 2 and rate 1, the ratio is given as; rate 2/ rate 1 = 12

Therefore the rate has increased by a factor of 12.

Answer:

An addiction could occur, maybe an overdose?, this could lead to death and maybe you would do unreasonable things which could get you fined or arrested.

Explanation:

Answer:

Hope this helped,

Kavitha

Answer:

For carbon, the effective nuclear charge is 3.25 and the valence electrons will reside in the orbitals 2s^2 and 2p^2

For silicon, the effective nuclear charge is 4.15 and its valance electrons will reside in the orbitals 3s^2 and 3p^2

Explanation:

Carbon

The effective nuclear charge of carbon is 3.25

To get the orbitals in which it’s valence electron reside, let’s write the electronic configuration

The atomic number of carbon is 6

So the configuration will be;

1s^2 2s^2 2p^2

So the valence electrons will reside in the orbitals 2s^2 and 2p^2

For silicon;

It’s effective nuclear charge is +4.15

The electronic configuration of silicon with atomic number 14 is;

1s^2 2s^2 2p^6 3s^2 3p^2

So the valence electrons will reside in the orbitals 3s^2 and 3p^2

Answer:

1. EF = PSCl₃; 2. MF = PSCl₃  

Explanation:

1. Empirical formula

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our first job is to calculate the molar ratio of P:S:Cl.

Assume 100 g of the compound.

(a) Calculate the mass of each element.

Then we have 18.28 g P, 18.93 g S, and 67.28 g Cl.

(b) Calculate the moles of each element

[tex]\text{Moles of P} = \text{18.28 g C} \times \dfrac{\text{1 mol P}}{\text{30.97 g P}} = \text{0.5902 mol P}\\\\\text{Moles of S} = \text{18.93 g S} \times \dfrac{\text{1 mol S}}{\text{32.06 g S }} = \text{0.5905 mol S}\\\\\text{Moles of Cl} = \text{62.78 g Cl} \times \dfrac{\text{1 mol Cl}}{\text{35.45 g Cl }} = \text{1.771 mol Cl}[/tex]

(c) Calculate the molar ratio of the elements

Divide each number by the smallest number of moles

P:S:Cl = 0.5902:0.5905:1.898 = 1:1.000:3.000 ≈ 1:1:3

(d) Write the empirical formula

EF = PSCl₃

The empirical formula for this compound is PSCl₃.

2. Molecular formula

(a) Calculate the ratio of the molecular and empirical formula masses

n = (169.4 u)/(169.40 u) = 1.000 ≈ 1

(b) Calculate the molecular formula

MF = (EF)ₙ = (EF)₁ = PSCl₃

The molecular formula for this compound is PSCl₃.

Answer:

THE HEAT OF COMBUSTION IS 4995.69 kJ/mol OF OCTANE.

Explanation:

Heat capacity = 6.18 kJ/C

Temperature change = 41.5 C - 22.0 C = 19.5 C

Heat required to raise the temperature by 19.5 °C is:

Heat = heat capacity * temperature change

Heat = 6.18 kJ/ C * 19.5 C

heat = 120.51 kJ of heat

120.51 kJ of heat is required to raise the temperature of 2.75 g sample of  a liquid octane.

Molar mass of octane = ( 12* 8 + 1 * 18) = 114 g/mol

So therefore, the heat of the reaction per mole of octane will be:

120.51 kJ of heat is required for 2.75 g of octane

x J of heat will be required for 114 g of octane

x J = 120.51kJ * 114 / 2.75

x = 4995.69 kJ of heat per mole.

In conclusion, the heat of the combustion reaction in kJ / mole of octane is 4995.69 kJ/mol

Complete question:

Write the condensed formula from left to right, starting with (CH3)x where x is a number.

See attached image for the structure formula of the compound

Answer:

(CH₃)₂CHC(CH₃)₃ named as 2,2,3-Trimethylbutane

Explanation:

If we number the longest chain of the carbon starting from the left, we will observe that there are four carbons in the straight chain as shown in the image.

Starting from first carbon from the left of the carbon chain, at carbon number number 2, there two alkyl group, that is two methyl (CH3 is two). Also at carbon number 3, there are three alkyl group, that is three methyl (CH3 is three).

The condensed formula will be written as;

(CH₃)₂CHC(CH₃)₃

This compound is named as 2,2,3-Trimethylbutane, an isomer of Heptane

Answer:

Poop Butt.

Explanation: Poop Butt.

Answer:

2PO₄³⁻ + 3Fe²⁺ → Fe₃(PO₄)₂(s)

Explanation:

In a net ionic equation you list only the ions that are participating in the reaction.

When potassium phosphate, K₃PO₄, reacts with iron (II) nitrate, Fe(NO₃)₂ producing iron (II) phosphate, Fe₃(PO₄)₂ that is an insoluble salt. The reaction is:

2K₃PO₄ + 3 Fe(NO₃)₂ → Fe₃(PO₄)₂(s) + 6NO₃⁻ + 6K⁺

The ionic equation is:

6K⁺ + 2PO₄³⁻ + 3Fe²⁺ + 6NO₃⁻→ Fe₃(PO₄)₂(s) + 6NO₃⁻ + 6K⁺

Subtracting the K⁺ and NO₃⁻ ions that are not participating in the reaction, the net ionic equation is:

Answer:

Explanation:

Given that:

mass of the antacid tablet = 5.8400 g

required mass of the antacid tablet = 4.2800 g was added to 200. mL of simulated stomach acid.

The amount of the  original 200. mL of stomach acid (in mL) needed to  neutralize the 4.2800 g crushed sample of the tablet can be calculated as:

= 11.6 mL of NaOH × 25.00 mL /29.0 mL NaOH

= 10.00 mL original stomach acid

Now; since it requires 11.6  mL of  NaOH o neutralize 10.00 mL of  original acid , then:

the antacid neutralized = 200 mL - 10.00 mL

the antacid neutralized = 190.00 mL

Answer:

As little as 2 days

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Answer: 44.37 degrees C

Explanation:

Use combined gas law: (P1)(V1)/T1=(P2)(V2)/T2

For most gas laws, you  must convert to Kelvin:

K=deg C+273

K=25+273=298 K

Plug and chug:

(1.0 atm)(1.2 L)/(298 K)=(0.71 atm)(1.8 L)/(x)

Solve for x and get 317.37 K

Subtract 273 from this to convert to degrees Celsius. You will get 44.37 degrees Celsius.

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Answer:

The balanced equation is: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)

Explanation:

Zn(s) is a simple substance (its oxidation number is zero) and it is oxidized to Zn²⁺. It loses two electrons, so the half reaction is the following:

Zn(s) → Zn²⁺(aq) + 2 e-   (oxidation reaction)

Hydrogen ion (H⁺) is reduced to hydrogen gas (H₂). The oxidation number is decreased from +1 to 0 (because H₂ is a simple substance). H⁺ gains 1 electron per H atom, so the half reaction is the following:

2H⁺(aq) + 2 e- → H₂(g) (reduction reaction)

We obtain the overall reaction from the addition of the two half reactions. We write the reduction reaction first and then the oxidation reaction, as follows:

2H⁺(aq) + 2 e- → H₂(g)

+

Zn(s) → Zn²⁺(aq) + 2 e-

---------------------------------

Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)

The two electrons at both sides of the equation (2 e-) are canceled. The overall reaction is in acidic solution due to the presence of H⁺ ions. The net charge at both sides is the same : +2, so the mass and the charge are balanced.

Answer:

ESCALAS MAYORES (D, E, G, A, B) Porfavor necesito ayuda,te lo agradecería muchísimo!!

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Answer:

The correct answer is statement A.

Explanation:

In a media comprising 20 percent ethyl acetate/hexane, as the benzophenone is non-polar, so it will travel farther with high Rf value. On the other hand, as benzohydrol is a polar molecule, therefore, it should be at lower Rf value and will not rise in the given media.  

For an experiment to be successful, there should not be any unreacted benzohydrol to be left in the experimental system. The benzophenone in the reaction as well as the standard samples should exhibit similar Rf value.  

The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value to analyze by TLC.

In synthetic chemistry, thin-layer chromatography (TLC) is a method that is frequently used to identify compounds, assess their purity, and monitor the progress of a reaction. Additionally, it enables the solvent system for a specific separation problem to be optimized.

The foundation of TCL is the adsorption-based separation theory. The separation depends on how sensitive different chemicals are to the stationary and mobile phases.

TLC, or thin layer chromatography, is a technique for isolating the components of mixtures before analysis. TLC can be used to identify compounds, ascertain their identities, and ascertain the purity of a compound.

Thus, option A is correct.

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:Answer : The limiting reactant is  and the theoretical yield of methanol is, 0.96 grams.

Explanation :

First we have to calculate the moles of  and .

where,

= pressure of CO gas = 232 mmHg = 0.305 atm   (1 atm = 760 mmHg)

V = volume of gas = 1.65 L

T = temperature of gas = 305 K

= number of moles of CO gas = ?

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

and,

where,

= pressure of  gas = 374 mmHg = 0.492 atm   (1 atm = 760 mmHg)

V = volume of gas = 1.65 L

T = temperature of gas = 305 K

= number of moles of  gas = ?

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 2 mole of  react with 1 mole of  

So, 0.0601 moles of  react with  moles 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.

Now we have to calculate the moles of  

From the reaction, we conclude that

As, 2 mole of  react to give 1 mole of  

So, 0.0601 moles of  react with  moles of  

Now we have to calculate the mass of  

Therefore, the theoretical yield of methanol is, 0.96 grams.

The theoretical yield of methanol is 0.496 g of methanol.

The reaction equation is CO (g) + 2H2 (g) → CH3OH (g).

From the partial pressures of each reactant, we can obtain the number of moles of reactants.

For CO;

P = 232 mmHg or 0.305 atm

V = 1.5 L

T = 305 K

n = ?

R = 0.082 atmL-1mol-1K-1

PV = nRT

n = PV/RT

n = 0.305 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K

n = 0.018 moles

For hydrogen;

P = 397 mmHg or 0.522 atm

V = 1.5 L

T = 305 K

n = ?

R = 0.082 atmL-1mol-1K-1

PV = nRT

n = PV/RT

n = 0.522 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K

n = 0.031 moles

From the reaction equation;

1 mole of CO reacted with 2 moles of H2

0.018 moles of CO will react with 0.018 moles × 2 moles/1 mole

= 0.036 moles of H2

We can see that there is not enough H2 to react with CO hence H2 is the limiting reactant.

2 moles of H2 yields 1 mole of methanol

0.031 moles of H2 yields  0.031 moles × 1 moles/2 mole

= 0.0155 moles of methanol

Mass of methanol produced = 0.0155 moles of methanol × 32 g/mol

= 0.496 g of methanol

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Answer:

A

Explanation:

If we intend to achieve the anti addition of Hex-3-yne to yield (E) Hex-3-ene, the we must use Na/NH3. The first step of the reaction involves the transfer of an electron from sodium to the alkene; this yields a radical anion. Strong electron replusion ensues between the single electron and the lone pair on the carbon. This now forces the both to be found at a trans position to each other and this is the basis of the stereochemistry of the product.

Secondly, the radical anion abstracts a proton from ammonia. Another sodium atom transfers an electron leading to the formation of a vinyl carbanion, the alkyl groups are now trans to each other.

This carbanion now abstracts a proton from ammonia and the final product is formed.

Answer:

Vapor pressure of water = 23.14torr

Explanation:

When you made a solution, vapor pressure decreases following Raoult's law:

[tex]P_{solution} = X_{solvent} P_{solvent}[/tex]

Where P is vapor pressure and X mole fraction

As vapor pressure of water is 23.77torr we must find the mole fraction of water knowing the solution is 1.500m glucose (That is 1.500 moles of glucose per kg of water = 1000g of water).

1000g of H₂O are, in moles (Molar mass: 18.02g/mol):

1000g H₂O ₓ (1mole / 18.02g) = 55.5 moles of H₂O.

As we know now the solution contains 55.5 moles of water and 1.5 moles of glucose. Thus, mole fraction of water (Solvent) is:

[tex]X_{H_2O} = \frac{55.5molesH_2O}{55.5molesH_2O + 1.5 molesGlucose} = 0.9737[/tex]

Replacing in Raoult's law, pressure of water above the solution is:

[tex]P_{solution} = X_{solvent} P_{solvent}[/tex]

[tex]P_{solution} = 0.9737*23.77torr[/tex]

Answer:

Most stable: 2,3-dimethylpent-2-ene > (E)-3,4-dimethylpent-2-ene > (Z)-3,4-dimethylpent-2-ene > 2-methyl-3-methylenepentane : Least stable

Explanation:

Treatment of NaOH with (R)-3-bromo-2,3-dimethylpentane results in the elimination of HBr. Each H atoms present on each [tex]\beta[/tex]-carbon atoms can be eliminated result in the formation of four possible products: (1) 2,3-dimethylpent-2-ene, (2) (E)-3,4-dimethylpent-2-ene, (3) (Z)-3,4-dimethylpent-2-ene and (4) 2-methyl-3-methylenepentane.

The stability of these alkenes depends on the number of hyperconjugative H atoms present with respect to the double bond. In accordance with this, 2,3-dimethylpent-2-ene is the most stable alkene (11-hyperconjugative H atoms). Then, 3,4-dimethylpent-2-ene is the second most stable alkene (7-hyperconjugative H atoms). Among (E)-3,4-dimethylpent-2-ene and (Z)-3,4-dimethylpent-2-ene, (E)-3,4-dimethylpent-2-ene is more stable due to it's less sterically hindered structure. 2-methyl-3-methylenepentane is the least stable alkene (3-hyperconjugative H atoms).

So, decreasing order of stability of alkenes from most stable to least stable:

2,3-dimethylpent-2-ene > (E)-3,4-dimethylpent-2-ene > (Z)-3,4-dimethylpent-2-ene > 2-methyl-3-methylenepentane

Three bromo As just a result of the nucleophilic substitution mechanism, 2,4 dimethylpentane generates a racemic mix containing both the r and s forms of the molecule.

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Answer:

0.5 M

Explanation:

First, let us look at the balanced equation of the reaction.

[tex]H_2SO_4 + 2KOH --> K_2SO_4 + 2H_2O[/tex]

The solute formed is [tex]K_2SO_4[/tex].

Recall that: mole = molarity x volume

Hence,

50 ml, 1.00 M H2SO4 = 0.05 x 1 = 0.05 mole

50 ml, 2.0 M KOH = 0.05 x 2 = 0.1 mole

From the equation

1 mole of H2SO4 reacts with 2 moles of KOH to give 1 mole of K2SO4.

Hence,

0.05 mole H2SO4 reacting with 0.1 mole KOH will give 0.05 mole [tex]K_2SO_4[/tex].

Also recall that: concentration = mole/volume

Total volume of resulting solution = 50 ml + 50 ml = 100 ml or 0.1 liter

Concentration of [tex]K_2SO_4[/tex] = mole of [tex]K_2SO_4[/tex]/volume of resulting solution

                              = 0.05/0.1 = 0.5 M

The concentration of the resulting solute = 0.5 M

Answer:

Amount of excess Carbon (ii) oxide left over = 23.75 g

Explanation:

Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂

Molar mass of Fe₂O₃ = 160 g/mol;

Molar mass of Carbon (ii) oxide = 28 g/mol

From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g)  of carbon (ii) oxide

450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide

Therefore the excess reactant is carbon (ii) oxide.

Amount of excess Carbon (ii) oxide left over = 260 - 236.25

Amount of excess Carbon (ii) oxide left over = 23.75 g

Answer:

Oxygen is oxidized and hydrogen is reduced

Explanation:

In the electrolysis of water, a pair of Platinum electrodes are immersed in water. Th water has a small quantity of either an acid, salt or base, in most cases H2S04, added to it, to aid ionization. This is because water on its own does not posses enough ions to undergo electrolysis. At the platinum anode, water is oxidized to oxygen gas and hydrogen ions. At the platinum cathode, water is reduced to hydrogen gas and hydroxide ions.  The proportion of oxygen and hydrogen produced should be theoretically 1 : 2 respectively, but is not usually so, due to competing side reactions. The hydrogen by product is usually used as  a fuel source, and it usually combine with the hydroxide ion to form water back again.

Radiologists are medical doctors that treat injuries using medical imaging (radiology)

Answer:

a person who uses X-rays or other high-energy radiation, especially a doctor specializing in radiology.

Explanation:

Answer:

F

Explanation:

Halogens may interact with the benzene ring via inductive or resonance effects. Halogens deactivate the benzene ring by inductive effect rather than by resonance effects.

The lone pairs of electrons present on the halogen atoms may be donated to the ring by resonance, but an opposite effect, the inductive pull (-I inductive effect) of the halogen atoms on electrons away from the benzene ring due to the high electro negativity of the halogens leads to a deactivation of the ring towards electrophilic substitution.

Hence inductive electron withdrawal by the halogen atom predominates over electron donation by resonance effect and the benzene ring g is deactivated towards electrophilic substitution at the ortho and para positions.

Answer:

The Nuclear magnetic resonance is the process this technique does not use radiation.

The  ms is an sensitive technology can be a massive number and small sample of the blood.

Explanation:

The Nuclear magnetic resonance we look at the both side of that coin.

The technique provides that fatty acid composition and various including amino acids.

These are contain the complementary these biomarkers, that are suitable for all kinds of studies. there are many types of research:-

(1) A powerful tool metabolic (2) A versatile tool research (3) Quick analysis (4) Low cost analysis.

The MS is an extremely sensitive technology using a very small number of the blood.

(1) Powerful techniques (2) Highly method (3) Large number of metabolites (4)Small sample volume

MS can be fine mapping metabolic pathways to sign analytical strategy.

Answer:

16.4 grams of calcium nitrate can be prepared by the reaction of 18.9 grams of nitric acid with 7.4 grams of calcium hydroxide

Explanation:

The balanced reaction is:

Ca(OH)₂ + 2 HNO₃ → Ca(NO₃)₂ + 2 H₂O

First, you must determine the limiting reagent. The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction):

Being:

Then, the molar mass of the compounds participating in the reaction is:

Then, by stoichiometry of the reaction the following amounts of reagents and products participate:

Then apply the following rule of three: if 126 grams of nitric acid reacts with 74 grams of calcium hydroxide, 18.9 grams of nitric acid with how much mass of calcium hydroxide does it react?

[tex]mass of calcium hydroxide=\frac{18.9 grams of nitric acid*74 grams of calcium hydroxide}{126 grams of nitric acid}[/tex]

mass of calcium hydroxide= 11.1 grams

But 11.1 grams of calcium hydroxide are not available, 7.4 grams are available. Since you have less mass than you need to react with 18.9 grams of nitric acid, calcium hydroxide will be the limiting reagent.

Then, it is possible to determine the amount of mass of calcium nitrate produced by another rule of three:  if 164 grams of calcium nitrate are formed by stoichiometry from 74 grams of calcium hydroxide, how much mass of calcium nitrate will form from 7.4 grams of calcium hydroxide?

[tex]mass of calcium nitrate=\frac{7.4 grams of calcium hydroxide*164 grams of calcium nitrate}{74 grams of calcium hydroxide}[/tex]

mass of calcium nitrate= 16.4 grams

16.4 grams of calcium nitrate can be prepared by the reaction of 18.9 grams of nitric acid with 7.4 grams of calcium hydroxide