Answer:
An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9 °C How much ice at a temperature of −17.5 °C must be dropped into the water so that the final temperature of the system will be 31.0 °C? Take the specific heat for liquid water to be 4190J/Kg.K, the specific heat for ice to be 2100J/Kg.K, and the heat of fusion for water to be 334000J/kg.
The answer to the above question is
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
Explanation:
To solve this we proceed by finding the heat reaquired to raise the temperature of the water to 31.0 C from 79.9 C then we use tht to calculate for the mass of ice as follows
ΔH = m×c×ΔT
= 0.205×4190×(79.9 -31.0) = 42002.655 J
Therefore fore the ice, we have
Total heat = mi×L + mi×ci×ΔTi = mi×334000 + mi × 2100 × (0 -−17.5) = 42002.655 J
370750×mi = 42002.655 J
or mi = 0.1133 kg
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
The abiotic factor that is quite used to support a relatively large amount of biotic factors is liquid(mostly water).
Answer:
Avogadro's number
Explanation:
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
It means 1 mole of any substance contain 6.022 × 10²³ number of representative particles.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
35.45 g Cl⁻ = 1mole = 6.022 × 10²³ Cl⁻ ions
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
Since that E°cell =E°cathode - E°anode then
0.62V = 0.34 - E°anode
Therefore E° anode is = -0.28V
