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2 years ago

Why can't liquids change volume but gasses can?

2 answers:

5 0

Gases don't have a definite shape or volume and fill any container. and since the particles are far apart from each other their attraction is weak. since there is a lots of free space between particles, gases can easily be compressed. water particles are held very close together by strong forces of attraction."Because the particles are close together, liquids can't easily be compressed and keep the same volume."

hope this helped

kupik [55]2 years ago

4 0

Because liquids cant be condensed the way that gasses can for example in a tank of argon you can put 20 cubic feet because it can be be condensed but you could not fit 20 cubic feet of water because it can not be packed together .

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Sulfonation of benzene has the following mechanism: (1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3− [slow]

ziro4ka [17]

Question is incomplete, complete question is as follows :

Complete Question : .Sulfonation of benzene has the following mechanism:

(1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3

[fast]

(2) SO3 + C6H6 → H(C6H5+)SO3−

[slow]

(3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4

[fast]

(4) C6H5SO3− + H3O+ → C6H5SO3H + H2O

[fast]

write the overall rate law for the initial rate of the reaction as a fraction.

Rate=k(________/_________)

Answer:

The overall rate law for the initial reaction is = $k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]$

Explanation :

Frist of all, all the common terms are cancelled out and written the overall reaction.

As we know that the rate depednant step is the slowest step of the reaction, rate law is :

rate = $k_{2} [SO_{3}][C_{6}H_{6}]$

But the problem is that SO3 cannot be written in the overall rate law because it is an intermediate.

Rate law for synthesis of S03 is as follows :

rate = $k_{1}[H_{2}SO_{4}]^{2}$

Hence when we substitute equation 2 in equation one,

Rate comes out to be = $k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]$

6 0

2 years ago

When 1.00 g of boron is burned in o2(g) to form b2o3(s), enough heat is generated to raise the temperature of 733 g of water fro

Bas_tet [7]

Answer: For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees. 4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ. Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work. To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3. .0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.

7 0

3 years ago

Suppose you carry out a titration involving 2.50 molar NaOH and an unknown concentration of HBr. To bring the reaction to its en

givi [52]

Answer:

Explanation:

Brainly Patrol ima stop you here

7 0

2 years ago

Bleach is a 0.750 M solution of sodium hypochlorite. What volume, in litres, would contain 120.3 g of sodium hypochlorite?

Lady bird [3.3K]

Answer:

2.15 L

Explanation:

M(NaOCl) = 23.0 + 16.0 + 35.5 = 74.5 g/mol

120.3 g *1 mol/74.5 g = 1.615 mol NaOCl

0.750 M = 0.750 mol/L

1.615 mol * 1L/0.750 mol = 2.15 L

4 0

3 years ago

One of the goal's of an atom is:

iren [92.7K]

Answer:

to have a "full" last orbital shell

Explanation:

They can not live withought ther outer most shell

5 0

2 years ago