I am sorry bro I don't know
Gr terlarut = 36 gr
<span>Mr terlarut = 180 </span>
<span>gr pelarut = 250 gr </span>
<span>Kb air = 0,52 °C kg/mol </span>
<span>Tb larutan = ........? </span>
<span>--------------------------------------... </span>
<span>ΔTb = Kb.m.i </span>
<span>ΔTb = Kb. (gr t / Mr t) . (1000/ gr p) .i </span>
<span>ΔTb = 0,52 x (36/180) x (1000/250) x 1 </span>
<span>ΔTb = 0,416 °C </span>
<span>Tb = 100 + ΔTb </span>
<span>Tb = 100 + 0,416 </span>
<span>Tb = 100,416 °C</span>
It's a conductor meaning metal is good for radiating heat.
The molar concentration of the KI_3 solution is 0.0833 mol/L.
<em>Step 1</em>. Calculate the <em>moles of S_2O_3^(2-)</em>
Moles of S_2O_3^(2-) = 25.00 mL S_2O_3^(2-) ×[0.200 mmol S_2O_3^(2-)/(1 mL S_2O_3^(2-)] = 5.000 mmol S_2O_3^(2-)
<em>Step 2</em>. Calculate the <em>moles of I_3^(-)
</em>
Moles of I_3^(-) = 5.000 mmol S_2O_3^(2-)))) × [1 mmol I_3^(-)/(2 mmol S_2O_3^(2-)] = 2.500 mmol I_3^(-)
<em>Step 3</em>. Calculate the <em>molar concentration of the I_3^(-)</em>
<em>c</em> = "moles"/"litres" = 2.500 mmol/30.00 mL = 0.083 33 mol/L
