Butter won't melt in a fridge because of intermolecular tensions. While the bonds inside of the fat molecules are unbroken, the attractions between the fat molecules are weaker.
What intermolecular forces are present in butter?
The intermolecular forces known as London dispersion forces are the weakest and are most prominent in hydrocarbons. Due to the fact that butter molecules are hydrocarbons, London dispersion forces do exist between them.
How do intermolecular forces affect melting?
More energy is required to stop the attraction between these molecules as the intermolecular forces become more powerful. Because of this, rising intermolecular forces are accompanied with rising melting points.
Which forces are intramolecular and which are intermolecular?
Intramolecular forces are those that hold atoms together within molecules. The forces that hold molecules together are known as intermolecular forces.
Learn more about intermolecular forces: brainly.com/question/9328418
#SPJ4
The molar mass of the gas is 77.20 gm/mole.
Explanation:
The data given is:
P = 3.29 atm, V= 4.60 L T= 375 K mass of the gas = 37.96 grams
Using the ideal Gas Law will give the number of moles of the gas. The formula is
PV= nRT (where R = Universal Gas Constant 0.08206 L.atm/ K mole
Also number of moles is not given so applying the formula
n= mass ÷ molar mass of one mole of the gas.
n = m ÷ x ( x molar mass) ( m mass given)
Now putting the values in Ideal Gas Law equation
PV = m ÷ x RT
3.29 × 4.60 = 37.96/x × 0.08206 × 375
15.134 = 1168.1241 ÷ x
15.134x = 1168.1241
x = 1168.1241 ÷ 15.13
x = 77.20 gm/mol
If all the units in the formula are put will get cancel only grams/mole will be there. Molecular weight is given by gm/mole.
Answer:
Q = 8.8 kJ
Explanation:
Step 1: Data given
The specific heat of a solution = 4.18 J/g°C
Volume = 296 mL
Density = 1.03 g/mL
The temperature increases with 6.9 °C
Step 2: Calculate the mass of the solution
mass = density * volume
mass = 1.03 g/mL * 296 mL
mass = 304.88 grams
Step 3: Calculate the heat
Q = m*c*ΔT
⇒ with Q = the heat in Joules = TO BE DETERMINED
⇒ with m = the mass of the solution = 304.88 grams
⇒ with c = the specific heat of the solution = 4.18 J/g°C
⇒ with ΔT = the change in temperature = 6.9 °C
Q = 304.88 g * 4.18 J/g°c * 6.9 °C
Q = 8793.3 J = 8.8 kJ
Q = 8.8 kJ
Answer:
1,31÷2 =10,11
Explanation:
c10h22+31÷2o2=10co2+11h2o
Answer:
Density (ρ) = 5 kilogram/cubic meter
Explanation:
Steps:
ρ =
m
V
=
10 kilogram
2 cubic meter
= 5 kilogram/cubic meter
