how does the volume of an ideal gas at constant temperature and pressure change as the number of moleculess increeases

Answers

Answer 1
Answer:

Answer:

Explanation:

This question seeks to test the knowledge of Avogadro's law. Avogadro's law states that equal volume of gases, at constant temperature and pressure,  contain the same number of molecules. The meaning here is that the volume of a given mass of (ideal) gas is directly proportional to the number of molecules. Thus, an increase in volume of an ideal gas will lead to an increase in the number of molecules of the gas. Also, an increase  in the number of molecules of an ideal gas will lead to an increase in the volume of the gas at constant temperature and pressure.


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In the first step, 4-sulfanilic acid reacts with sodium nitrate to form diazonium ion intermediate. Identify the Lewis acid and Lewis base in this reaction.
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What is an extensive property that can be calculated?

50.0ml each of 1.0M Hcl and 1.0M Naoh at room temperature (20.0c) are mixed the temperature of the resulting Nacl solutions increase to 27.5cthe density if the resulting Nacl solutuion 1.02 g/ml
the specific heat of the resulting Nacl solutions is 4.06j/gc
calculate the heat of neutralisation of hcl and naoh in kj/mol nacl products​

Answers

Answer:

62.12kJ/mol

Explanation:

The neutralization reaction of HCl and NaOH is:

HCl + NaOH → NaCl + H₂O + HEAT

You can find the released heat of the reaction and heat of neutralization (Released heat per mole of reaction) using the formula:

Q = C×m×ΔT

Where Q is heat, C specific heat of the solution (4.06J/gºC), m its mass and ΔT change in temperature (27.5ºC-20.0ºC = 7.5ºC).

The mass of the solution can be finded with the volume of the solution (50.0mL of HCl solution + 50.0mL of NaOH solution = 100.0mL) and its density (1.02g/mL), thus:

100.0mL × (1.02g / mL) = 102g of solution.

Replacing, heat produced in the reaction was:

Q = C×m×ΔT

Q = 4.06J/gºC×102g×7.5ºC

Q = 3106J = 3.106kJ of heat are released.

There are 50.0mL ×1M = 50.0mmoles = 0.0500 moles of HCl and NaOH that are reacting releasing 3.106kJ of heat. That means heat of neutralization is:

3.106kJ / 0.0500mol of reaction =

62.12kJ/mol is heat of neutralization

Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-chlorobutane. Based on this information, the major organic product(s) of this reaction are expected to be _____.

Answers

Answer:but-1-ene

Explanation:This is an E2 elimination reaction .

Kindly refer the attachment for complete reaction and products.

Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.

Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .

As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.

Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.

2-butene is more thermodynamically6 stable as compared to 1-butene  

The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.

A car is traveling at 87.0 km/hr. How many meters will it travel in 37.0 seconds?

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Speed of the car = 87.0 (km)/(hr)

Given time = 37.0 s

Converting time from seconds to hours:

1 hr= 60 min; 1 min = 60 s

37.0 s * (1 min)/(60s)*(1 hr)/(60min)  = 0.0103 hr

Calculating distance from speed and time:

0.0103 hr * (87.0 km)/(1 hr) =0.894 km

Converting distance from km to m:

0.894 km * (1000 m)/(1 km) = 894 m

So the distance traveled by the car in 37.0 s is 894 m.

15. List the substances A-E in order from most dense to least dense based on the facts provided.Please write the letters in the spaces provided below.
Substance A: 8.2 g/cm3

Substance B: 3.5 cm and 30.0g

Substance C: 10.0g and 40mL

Substance D: 0.5 g/cm3

Substance E: 2.0cm by 3.0cm by 1.0cm and 4.0g

Most Dense_ _ _ _ _
Least Dense

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The order of density of substances ranging from most dense to least dense is :substance B>substance A>substance E>substance C>substance D.

What is density?

It is a ratio of substance's mass per unit of volume.Symbol most commonly used for density is р.The SI unit of densityis kilogram per cubic meter .It explains how tightly a material is packed together.

There are2 types of density :1)absolute density 2) relativedensity.Absolute density is the massof any  substance per unit volume and relative density is the ratio of density of a substance to the density of a given reference material.

Reference material used forrelative density is water.The instrument used for measuring density or relative density of liquids is hydrometer. Densityis measured at constant temperature and pressure.

To learn more about density and it's types click here:

brainly.com/question/15164682

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Answer:

So 1st it is B then D then E then a then C

Which is the most stable? Carbon (C) Sodium (Na) Helium (He)

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the answer is helium because

How much is a share in namsek worth as a percent of a share in oxd group

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Answer:

10% is a share in NAMSEK worth as a percentage of a share in ODX Group Inc., in year 4

EXPLANATION:give brainliest

The share of NAMSEK in year 4 is around $  shares (as given from the above table). Hence the value of one share will be $2.5/share

The share of ODX Group Inc., in year 4 is around $ shares (as given from the above table). Hence the value of one share is $

Therefore a share in NAMSEK worth as a percentage of a share in ODX Group Inc., in year 4 will be  

= 0.1 or 10 %