When one mole of Na3PO4.3H2O is heated extensively, three moles of water are released.
The water molecules in Na3PO4.3H2O are called molecules of water of crystallization. These molecules are not covalently bonded to the Na3PO4 molecule. They are only loosely attached to the substance.
Strong heating will drive away these molecules of water of crystallization to give three moles of water in the product.
Hence, when one mole of Na3PO4.3H2O is heated extensively, three moles of water are released.
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1 mole of h3bo3...........6.023*10²³ each h and B and 0 so we will have 3hydrogen+ 1 B+3 oxygen = 7*6.023*10²³ atoms
1 mole .......7*6.023*10²³atoms
4 moles ........x atoms
x=4*7*6.023*10²³.
Answer:
0.70 J/g.°C
Explanation:
Step 1: Given data
- Mass of graphite (m): 402 g
- Heat absorbed (Q): 1136 J
- Initial temperature: 26°C
- Specific heat of graphite (c): ?
Step 2: Calculate the specific heat of graphite
We will use the following expression.
Q = c × m × ΔT
c = Q / m × ΔT
c = 1136 J / 402 g × (30°C - 26°C)
c = 0.70 J/g.°C
Explanation:
Apply the mass of balance as follows.
Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank



[/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = 

+ h = 1
= 1 - h
= dt
= t + C
Given at t = 0 and V = 0
= 0
or, h = 0
-ln(1 - h) = t + C
Initial condition is -ln(1) = 0 + C
C = 0
So, -ln(1 - h) = t
or, t =
........... (1)
(a) Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.
t =
t =
= 
= 0.916 seconds
(b) As maximum height of water level in the tank is achieved at steady state that is, t =
.
1 - h = exp (-t)
1 - h = 0
h = 1
Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.