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SOVA2 [1]
4 years ago
8

Which best explains why water is able to “stick” to the side of glass?

Chemistry
2 answers:
solong [7]4 years ago
7 0
The answer which best explains why water is able to "stick" to the side of glass is the following one - <span>Strong adhesive forces exist between glass and the water molecules.</span>
AveGali [126]4 years ago
4 0

Explanation:

Adhesion means the ability to stick on the surface of another substance.

Water exhibits adhesive forces due to which it is able to stick to the glass.  Due to adhesive forces water spreads over the surface of glass and sticks to it.

These adhesive forces between the glass and water enough that it deforms the spherical shape of water molecules and help them stick to the surface of glass. As a result, adhesive forces overcome the repulsion between like molecules.

Hence, water is able to “stick” to the side of glass due to strong adhesive forces.


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Neutrons are neutral, so they don’t have a positive or negative charge. Hence, the cashier says no charge for the neutron. It is quite funny when you think about it lol.
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Calculate the relative molecular mass of Ma(HCO3)2​
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Read 2 more answers
From the data below, calculate the total heat (in j) needed to convert 0.782 mol of gaseous ethanol at 300.0°c and 1 atm to liqu
Anika [276]

Answer:

You must remove \text{50.6 kJ} .

Explanation:

There are three heat transfers in this process:

Total heat = cool the vapour + condense the vapour + cool the liquid  

       q          =           q₁            +                q₂                   +           q₃

       q          =       nC₁ΔT₁        +          nΔHcond             +        nC₂ΔT₂

Let's calculate these heat transfers separately.

Data:

You don't give "the data below", so I will use my best estimates from the NIST Chemistry WebBook. You can later substitute your own values.

C₁ = specific heat capacity of vapour = 90 J·K⁻¹mol⁻¹

C₂ = specific heat capacity of liquid   = 115 J·K⁻¹mol⁻¹

ΔHcond = -38.56 kJ·mol⁻¹

Tmax = 300   °C

  b.p. =   78.4 °C

Tmin =   25.0 °C

n = 0.782 mol

Calculations:

ΔT₁ = 78.4 - 300 = -221.6 K

q₁ = 0.782 × 90 × (-221.6) = -15 600 J = -15.60 kJ

q₂ = 0.782 × (-38.56) = -30.15 kJ

ΔT = 25.0 - 78.4 = -53.4 K

q₃ = 0.782 × 115 × (-53.4) = -4802 J = 4.802 kJ

q = -15.60 - 53.4 - 4.802 = -50.6 kJ

You must remove \text{50.6 kJ} of heat to convert the vapour to a gas.

8 0
4 years ago
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