Answer:
Diagram Z
Explanation:
A cell placed into a hypotonic solution will swell and expand until it eventually burst through a process known as cytolysis.
Answer:
25.0 mL
Explanation:
1. Gather the information in one place.
MM: 98.00 74.09
2H3PO4 + 3Ca(OH)2 → Ca3(PO4)2 + 6H2O
m/g: 8.85 15.76
V/mL: 350.0 550
2. Moles of H3PO4
n = 8.85 g × (1 mol/98.00 g) = 0.09031 mol H3PO4
3. Moles of Ca(OH)2
n = 15.76 g × (1 mol/74.09 g) = 0.2126 mol Ca(OH)2
4. Moles of Ca(OH)2 in 25.0 mL Solution
n = 0.2126 mol × (25.0 mL/550 mL) = 0.009 663 mol Ca(OH)2
5. Moles of H3PO4 needed
From the balanced equation, the molar ratio is 2 mol H3PO4: 3 mol Ca(OH)2
n = 0.009 663 mol Ca(OH)2 × (2 mol H3PO4/3 mol Ca(OH)2)
= 0.006 442 mol H3PO4
6. Volume of H3PO4
V = 0.006 442 mol ×( 350.0 mL/0.09031 mol) = 25.0 mL H3PO4
It will take 25.0 mL of the H3PO4 solution to neutralize 25.0 mL of the Ca(OH)2 solution.
Answer: Enthalpy for the dissolution reaction of one mole of aluminum nitrate is -158 kJ/mol.
Explanation:
It is known that the density of water is 1 g/mL. So, mass of water will be calculated as follows.
Mass = volume × density
= 
= 103.2 g
Specific heat of water is 4.18 
.
Now, we will calculate the enthalpy of solution as follows.
= 
= 2372.5 J
As, 1 J = 
. So, 2372.5 J will be converted into kJ as follows.
= 2.37 kJ
Molar mass of 
= 213 g/mol
Hence, moles of will be calculated as follows.
Moles of = 
= 0.015
Therefore, enthalpy for the dissolution will be calculated as follows.
= -158 kJ/mol
Thus, we can conclude that enthalpy for the dissolution reaction of one mole of aluminum nitrate is -158 kJ/mol.
Answer:
the number of heat units needed to raise the temperature of a body by one degree.
