The new concentration of sodium phosphate is 0.315 mol·L⁻¹.
We can use the dilution formula
:
<em>c</em>₁<em>V</em>₁ = <em>c</em>₂<em>V</em>₂
We can rearrange the formula to get
<em>c</em>₂= <em>c</em>₁ × (<em>V</em>₁/<em>V</em>₂)
∴ <em>c</em>₂ = 1.50 mol·L⁻¹ × (21.00 mL)/(100.0 mL)] = 1.50 mol·L⁻¹ × 0.2100
= 0.315 mol·L⁻¹
Answer: The energy required to vaporize 12.5 g of liquid water is 28.2 kJ
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 gram of liquid into its vapor state without change in its temperature.
Given : The enthalpy of vaporization of water is 40.65 kJ/mol.
n = number of moles =
Thus 1 mole of water requires heat = 40.65 kJ
0.694 moles of water requires heat =
Thus the energy required to vaporize 12.5 g of liquid water is 28.2 kJ
When two atoms of the same element are covalently bonded, the radius of each atom will be half the distance between the two nuclei because they equally attract the electrons. The reason for this trend is that the bigger the radii, the further the distance between the two nuclei. Hope this helps:)