<span><span>m1</span>Δ<span>T1</span>+<span>m2</span>Δ<span>T2</span>=0</span>
<span><span>m1</span><span>(<span>Tf</span>l–l<span>T<span>∘1</span></span>)</span>+<span>m2</span><span>(<span>Tf</span>l–l<span>T<span>∘2</span></span>)</span>=0</span>
<span>50.0g×<span>(<span>Tf</span>l–l25.0 °C)</span>+23.0g×<span>(<span>Tf</span>l–l57.0 °C)</span>=0</span>
<span>50.0<span>Tf</span>−1250 °C+23.0<span>Tf</span> – 1311 °C=0</span>
<span>73.0<span>Tf</span>=2561 °C</span>
<span><span>Tf</span>=<span>2561 °C73.0</span>=<span>35.1 °C</span></span>
the answer you are looking for is 4.22 kj.
The correct answer is 0.5 kg If you whant me to explaine pls tell me
From the calculation as shpwn in the procedure below, the equilibrium constant of the substance is 6.9 * 10^-15.
<h3>What is equilibrium constant?</h3>
The equilibrium constant for the solubility of aa solid in solution is called the solubility product Ksp. The Ksp shows the extent to which a solid is dissolved in solution.
Given that;
Fe(OH)2 ⇄Fe^2+ + 2(OH)^-
Ksp = s(2s)^2
We have s as 1.2 x 10^-5 M
So
Ksp = 4s^3
Ksp = 4( 1.2 x 10^-5 )^3
Ksp = 6.9 * 10^-15
Learn more about Ksp:brainly.com/question/27132799
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