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>
Answer:
The pH of the lye in soap is 13.
Explanation:
Answer:
I just need points so ignore my answer and thanks
Answer:
The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year
Explanation:
Given that the mass of the carbon 14 at the start = 5 gram
At the end of 5,000 years we will have;

Where
A = The amount of carbon 14 left
A₀ = The starting amount of carbon 14
e = Constant = 2.71828
= The half life

t = The time elapsed = 5000 years
λ = 0.693/
= 0.693/5730 = 0.0001209424
Therefore;
A = 5 × e^(-0.0001209424×5000) = 2.7312 grams
Therefore, the amount of carbon 14 decayed in the 5000 years is the difference in mass between the starting amount and the amount left
The amount of carbon 14 decayed = 5 - 2.7312 = 2.2688 grams
The average yearly rate of change of carbon-14 during the first 5000 years is therefore;
2.2688 grams/(5000 years) = 0.0004538 grams per year
The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year.