<span>H2O water molecule is inorganic. Organic molecules contains carbon atom. Pentane, Butane, and Glucose have carbon in their molecules. Water is inorganic it contains hydrogen and oxygen. there is no carbon molecule. Carbohydrates, lipids etc are examples of organic molecules.</span>
Answer:
<em>293.99 g </em>
OR
<em>0.293 Kg</em>
Explanation:
Given data:
Lattice energy of Potassium nitrate (KNO3) = -163.8 kcal/mol
Heat of hydration of KNO3 = -155.5 kcal/mol
Heat to absorb by KNO3 = 101kJ
To find:
Mass of KNO3 to dissolve in water = ?
Solution:
Heat of solution = Hydration energy - Lattice energy
= -155.5 -(-163.8)
= 8.3 kcal/mol
We already know,
1 kcal/mol = 4.184 kJ/mole
Therefore,
= 4.184 kJ/mol x 8.3 kcal/mol
= 34.73 kJ/mol
Now, 34.73 kJ of heat is absorbed when 1 mole of KNO3 is dissolved in water.
For 101 kJ of heat would be
= 101/34.73
= 2.908 moles of KNO3
Molar mass of KNO3 = 101.1 g/mole
Mass of KNO3 = Molar mass x moles
= 101.1 g/mole x 2.908
= 293.99 g
= 0.293 kg
<em><u>293.99 g potassium nitrate has to dissolve in water to absorb 101 kJ of heat. </u></em>
Answer:
The mass of water 
= 39.18 gm
Explanation:
Mass of iron 
= 32.5 gm
Initial temperature of iron 
= 22.4°c = 295.4 K
Specific heat of iron 
= 0.448 
Mass of water =
Specific heat of water 
Initial temperature of water 
= 336 K
Final temperature after equilibrium 
= 59.7°c = 332.7 K
When iron rod is submerged into water then
Heat lost by water = Heat gain by iron rod
( - ) = ( - )
Put all the values in above formula we get
× 4.2 × ( 336 - 332.7 ) = 32.5 × 0.448 × ( 332.7 - 295.4 )
= 39.18 gm
Therefore the mass of water = 39.18 gm
Answer:
V2 = 3.11 x 105 liters
Explanation:
Initial Volume, V1 = 2.16 x 105 liters
Initial Temperature, T1 = 295 K
Final Temperature, T2 = 425 K
Final Volume, V2 = ?
These quantities are related by charle's law and the equation of the law is given as;
V1 / T1 = V2 / T2
V2 = T2 * V1 / T1
V2 = 425 * 2.16 x 105 / 295
V2 = 3.11 x 105 liters
Answer:
Explanation:
2 moles hydrogen reacts with one mole of oxygen to give 2 moles of water.
a ) rate of consumption of hydrogen ( moles per second) is twice the rate of consumption of oxygen .
b ) rate of formation of water ( moles per second ) is twice the rate of consumption of oxygen
c ) rate of formation of water ( moles per second ) is equal to the rate of consumption of hydrogen.
