Mass of NaCl = 750 g
mass of water = 1000 kg
Kf = - 1.86 °C
moles of NaCl =
=
= 12.8 moles
molality of NaCl =
=
Δ Tf = Kf * m = -1.86 * 0.01282 = - 0.024 °C
It has 1 valence electron.
Answer:
0.0898M is the molarity of the basic solution
Explanation:
The reaction of calcium hydroxide, Ca(OH)₂ with H₃PO₄ is:
3 Ca(OH)₂ + 2 H₃PO₄ → 6H₂O + Ca₃(PO₄)₂
To solve this question we must find the moles of H3PO4 that react. With the moles and the rection we can find the moles of Ca(OH)2. Using its volume we can find its molarity:
<em>Moles H3PO4:</em>
0.0189L * (0.200mol / L) = 0.00378 moles H3PO4
<em>Moles Ca(OH)2:</em>
0.00378 moles H3PO4 * (3mol Ca(OH)2 / 2mol H3PO4) = 0.00567 moles Ca(OH)2
<em>Molarity:</em>
0.00567 moles Ca(OH)2 / 0.06315L =
<h3>0.0898M is the molarity of the basic solution</h3>
Answer:
True
Explanation:
In an ionic reaction, the net ionic equation is used to calculate the standard enthalpy of reaction.
Hence the standard enthalpy of any ionic reaction can be obtained by considering its balanced net ionic reaction equation, hence the answer above.
