Answer:
0.00735°C
Explanation:
By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water
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<u>The</u><u> </u><u>elevation</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>is</u><u> </u><u>a</u><u> </u><u>phenomenon</u><u> </u><u>in</u><u> </u><u>which</u><u> </u><u>there</u><u> </u><u>is</u><u> </u><u>increase</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>in</u><u> </u><u>solution</u><u>,</u><u> </u><u>when</u><u> </u><u>the</u><u> </u><u>particular</u><u> </u><u>type</u><u> </u><u>of</u><u> </u><u>solute</u><u> </u><u>is</u><u> </u><u>added</u><u> </u><u>to</u><u> </u><u>pure</u><u> </u><u>solvent</u><u>.</u>
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Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.
and 'i' is 3 (as given in the question)
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'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)
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'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.
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<u>To</u><u> </u><u>find</u><u> </u><u>molality</u><u>,</u><u> </u><u>we</u><u> </u><u>have</u><u> </u><u>to</u><u> </u><u>divide</u><u> </u><u>no</u><u>.</u><u> </u><u>of</u><u> </u><u>moles</u><u> </u><u>of</u><u> </u><u>solute</u><u> </u><u>by</u><u> </u><u>weight</u><u> </u><u>of</u><u> </u><u>solution</u>
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While first we need to no. of moles

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<u>Now</u><u>,</u><u> </u><u>we</u><u> </u><u>will</u><u> </u><u>find</u><u> </u><u>molality</u>
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<u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>change</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>7</u><u>3</u><u>5</u><u>°</u><u>C</u><u>.</u>
POH = - log [ OH-]
pOH = - log [ 1 x 10⁻¹²]
pOH = 12
An organism that does not move on its own and makes food from its environment is a producer.
<span>We can use the heat
equation,
Q = mcΔT </span>
<span>Where Q is
the amount of energy transferred (J), m is the mass of the
substance (kg), c is the specific heat (J g</span>⁻¹ °C⁻<span>¹) and ΔT is the temperature
difference (°C).</span>
Density = mass / volume
The density of water = 0.997 g/mL
<span>Hence mass of 1.25 L (1250 mL) of water = 0.997 g/mL x 1250 mL</span>
<span> = 1246.25 g</span>
Specific heat capacity of water = 4.186 J<span>/ g °C.</span>
Let's assume that there is no heat loss to the surrounding and the final temperature is T.
By applying the equation,
5430 J = 1246.25 g x 4.186 J/ g °C x (T - 23) °C
(T - 23) °C = 5430 J / 1246.25 g x 4.186 J/ g °C
(T - 23) °C = 1.04 °C
T = 1.04 °C + 23 °C
T = 24.04 °C
Hence, the final temperature of the water is 24.04 °C.