Answer:
T2 =21.52°C
Explanation:
Given data:
Specific heat capacity of sample = 1.1 J/g.°C
Mass of sample = 385 g
Initial temperature = 19.5°C
Heat absorbed = 885 J
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )
885 J = 423.5 J/°C× (T2 - 19.5°C )
885 J / 423.5 J/°C = (T2 - 19.5°C )
2.02°C = (T2 - 19.5°C )
T2 = 2.02°C + 19.5°C
T2 =21.52°C
True!
Energy released by system is absorbed by surroundings.
<h2>Answer:</h2>
Arrangement of inter molecular forces from strongest to weakest.
- Hydrogen bonding
- Dipole-dipole interactions
- London dispersion forces.
<h3>Explanation:</h3>
Intermolecular forces are defined as the attractive forces between two molecules due to some polar sides of molecules. They can be between nonpolar molecules.
Hydrogen bonding is a type of dipole dipole interaction between the positive charge hydrogen ion and the slightly negative pole of a molecule. For example H---O bonding between water molecules.
Dipole dipole interactions are also attractive interactions between the slightly positive head of one molecule and the negative pole of other molecules.
But they are weaker than hydrogen bonding.
London dispersion forces are temporary interactions caused due to electronic dispersion in atoms of two molecules placed together. They are usually in nonpolar molecules like F2, I2. they are weakest interactions.
Answer:
No, in science their meanings are not the same as their everyday meanings.
Explanation:
In Science, Precision and Accuracy are defined as,
Accuracy:
Accuracy is the value which is closest to the known or standard value.
Precision:
While, Precision is the value of closeness of two measured values to each other.
Example:
Let suppose in Chemistry Lab you weight an object as 50 g. While the actual weight of that object is 30 g. It means your reading is not accurate.
On second measurement you find that the object weight is 31 g. This time your reading is not precise.
