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OlgaM077 [116]
3 years ago
14

what does it mean when the h3o+ to x- distance doesn't change much as the first 4 to 5 water molecules are added

Chemistry
2 answers:
podryga [215]3 years ago
7 0
If X is an equivalent base to H₂O
HX is an equivalent acid to H₃O⁺
HX is a stronger acid than H₃O⁺
HX is not an acid
X⁻ is a stronger base than H₂O
HX is a weaker acid than H₃O⁺
X⁻ is a weaker base than H₂O
X⁻ is not a base.

The correct response or this is 
X⁻ is a stronger base than H₂O
HX is a weaker acid than H₃O⁺
algol [13]3 years ago
6 0

Gjvdigcyivcghjnjjxhfbhdbdbbxbfbdbbbfbdbbdhdbbdbdbxbxbbcbcbbxbxbxbbxbdbbcbdbdbdbhdbfbcbbcbfbdbbcbbcbxbbxbcbcbbcbbxbxbbxbbxbxbbdbdbcbbdbbdbbxbbxbbxbbfbfbxbdbbdbdbdbbxbbfbbfbfbfbbfbfbccbbcbcbbcbxbxbbxbx

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True or False: All material has the same density.
Burka [1]

Answer:

Different materials have different densities. So it is False

4 0
3 years ago
Read 2 more answers
Ammonium nitrate dissociates in water according to the following equation:
Viktor [21]

Answer:

Explanation:

NH₄NO₃ = NH₄⁺ +NO₃⁻

heat released  by water = msΔ T

m is mass , s is specific heat and ΔT is fall in temperature

= 50  x 4.18 x ( 22 - 16.5 )  ( mass of 50 mL is 50 g )

= 1149.5 J .

This heat will be absorbed by the reaction above .

q for the reaction = + 1149.5 J

2 )

molecular weight of NH₄NO₃ = 80

No of moles reacted = 5/80 = 1 / 16 moles.

3 )  

5 g absorbs 1149.5 J

80 g absorbs 1149.5 x 16 J

= 18392 J

= 18.392 kJ.

= + 18.392 kJ

ΔH =  18.392 kJ / mol

6 0
3 years ago
Electrons are embedded in a mass of positively<br> charged matter.
grandymaker [24]

Answer:

plum pudding model .

Explanation:

the electrons were 'like plums embedded in a pudding'. Also called the Raisin Bread Model.

5 0
3 years ago
77 grams of an unknown metal at 99ᵒC is placed in 225 grams of water which is initially at 22ᵒC. The water is inside a 44 gram a
BigorU [14]

Answer:

The specific heat capacity of the unknown metal is C = 0.6991 J/g°C = 0.1671 cal/g°C

Explanation:

Heat lost by the unknown metal is equal to the heat gained by the water and aluminium cup.

Given,

Mass of unknown metal = 77 g

Initial Temperature of unknown metal = 99°C

Mass of water = 225 g

Initial Temperature of water = 22°C

Mass of Aluminium cup = 44 g

Specific heat capacity of Aluminium cup = 0.22 cal/gᵒC = 0.92048 J/g°C

Final temperature of the setup = 26°C

Note that, specific heat capacity of water = 4.186 J/g°C

Let the specific heat of the unknown metal be C

Heat lost from the unknown metal

= (77)(C)(99 - 26) = (5,621C) J

Heat gained by water

= (225)(4.186)(26 - 22) = 3,767.4 J

Heat gained by Aluminium cup

= (44)(0.92048)(26 - 22) = 162.00448 J

Heat lost by unknown metal = (Heat gained by water) + (Heat gained by Aluminium cup)

5621C = 3,767.4 + 162.00448 = 3,929.40448

5621C = 3,929.40448

C = (3,929.40448 ÷ 5621) = 0.6991 J/g°C

C = 0.6991 J/g°C = 0.1671 cal/g°C

Hope this Helps!!!

6 0
3 years ago
Read 2 more answers
A pond has the dimensions of 25 meters long by 20 meters wide. If there are 100 frogs living in the pond, what is the "populatio
sweet-ann [11.9K]

Answer:

D. 5 frogs/sq meter

Explanation:

5 0
3 years ago
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