Answer:
Heat required to melt 26.0 g of ice at its melting point is 8.66 kJ.
Explanation:
Number of moles of water in 26 g of water: 26×
moles
=1.44 moles
The enthalpy change for melting ice is called the entlaphy of fusion. Its value is 6.02 kj/mol.
we have relation as:
q = n × ΔH
where:
q = heat
n = moles
Δ
H = enthalpy
So calculating we get,
q= 1.44*6.02 kJ
q= 8.66 kJ
We require 8.66 kJ of energy to melt 26g of ice.
To cut this short and for your understanding, ionic bond is formed between metals (mostly right column in periodic table). Covalent bond is formed between non-metals (mostly left column in periodic table). So polar covalent is also a covalent bond but it is polar, which means the shape of molecules are not symmetrical hence maybe an atom in a molecule has most of the electron attracted to it causing itself to be partial negative (since electron are negatively charged) and the other atom has its electron being attracted by others became partial positive. Polar covalent can also be when H atom is binding either to F, O or N (also known as hydrogen bond).
Answer:
The right answer is "3 g".
Explanation:
Given:
Initial mass substance,

By using the relation between half lives and amount of substances will be:
⇒ 


Thus, the above is the correct answer.
Answer: 2800 calories
Explanation:
Latent heat of fusion is the amount of heat required to convert 1 mole of solid to liquid at atmospheric pressure.
Amount of heat required to fuse 1 gram of water = 80 cal
Mass of ice given = 35 gram
Heat required to fuse 1 g of ice at
= 80 cal
Thus Heat required to fuse 35 g of ice =
Thus 2800 calories of energy is required to melt 35 g ice cube
Answer:
v = 10 km/h
Explanation:
Step 1: Given data
- Distance traveled in the Bike Trip (d): 1 km
- Time elapsed in the Bike Trip (t): 0.1 h
Step 2: Calculate the speed in the Bike Trip
The speed (v) is equal to the distance traveled divided by the time elapsed. We will use the following mathematical expression.
v = d/t
v = 1 km/0.1 h
v = 10 km/h
The speed is 10 kilometers per hour.