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Ymorist [56]
3 years ago
5

Describe how plants might be affected if the inter molecular forces in water decreased? i need asap!

Chemistry
1 answer:
vovangra [49]3 years ago
6 0

Answer:

Water has polar O-H bonds. The negative O atoms attract the positive H atoms in nearby molecules, leading to the unusually strong type of dipole-dipole force called a hydrogen bond. Since water has hydrogen bonds, it also has dipole-induced dipole and London dispersion forces.

Hope it helped!!

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A 28.5 g piece of gold is heated and then allowed to cool. What is the change in temperature (°C) if the gold releases 0.106 kJ
Shkiper50 [21]

Considering the definition of calorimetry and sensible heat, the change in temperature if the gold releases 0.106 kJ of heat as it cools is 28.78°C.

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change). It is calculated by the expression:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

  • Q= 0.106 kJ= 106 J (being 1 kJ=1000 J)
  • c= 25.4 \frac{J}{molC}
  • m= 28.5 g×\frac{1 mole}{196.967 g} = 0.145 moles (being 196.967 g/mole the molar mass of gold)
  • ΔT= ?

Replacing:

106 J= 25.4 \frac{J}{molC}× 0.145 moles× ΔT

Solving:

ΔT=\frac{106 J}{25.4\frac{J}{molC}x0.145 moles}

ΔT= 28.78 C

The change in temperature if the gold releases 0.106 kJ of heat as it cools is 28.78°C.

Learn more:

  • brainly.com/question/11586486?referrer=searchResults
  • brainly.com/question/24724338?referrer=searchResults
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What is the expected freezing point of a 0.50 m solution of na2so4 in water kf for water is 1.86°c/m?
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Colligative properties calculations are used for this type of problem. Calculations are as follows:<span>


ΔT(freezing point)  = (Kf)m
ΔT(freezing point)  = 1.86 °C kg / mol (0.50 mol/kg)
ΔT(freezing point)  = 0.93 °C
Tf - T = 0.93 °C
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