4 moles of sugar.
Explanation:
A mole is defined as the amount of a substance contained in Avogadro's number of particles 6.02 x 10²³.
1 mole of substance = 6.02 x 10²³. molecules
Given that;
the sample of sugar contains 1.505 x 10²³.molecules
The number of moles in this amount of sugar is 4 moles
Learn more:
Number of moles brainly.com/question/13064292
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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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![\textsf {While} \: \sf {\Delta T_b} \: \textsf{expression is used} \\ \textsf {for elevation of boiling point}](https://tex.z-dn.net/?f=%20%5Ctextsf%20%7BWhile%7D%20%5C%3A%20%20%5Csf%20%20%7B%5CDelta%20T_b%7D%20%20%5C%3A%20%5Ctextsf%7Bexpression%20is%20used%7D%20%5C%5C%20%20%5Ctextsf%20%7Bfor%20elevation%20of%20boiling%20point%7D)
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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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![\sf \large \underline{The \: formula \: to \: be \: used \: in \: this \: question \: is} \\ \boxed{T_b = i \times K_b \times m}](https://tex.z-dn.net/?f=%20%5Csf%20%20%5Clarge%20%5Cunderline%7BThe%20%5C%3A%20%20formula%20%5C%3A%20to%20%5C%3A%20%20be%20%20%5C%3A%20used%20%5C%3A%20%20in%20%5C%3A%20%20this%20%5C%3A%20%20question%20%5C%3A%20%20is%7D%20%20%5C%5C%20%20%20%5Cboxed%7BT_b%20%3D%20i%20%5Ctimes%20%20K_b%20%5Ctimes%20%20m%7D)
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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
![\sf \implies no. \: of \: moles = \frac{weight \: of \: solute}{molar \: mass \: of \: solute} \\ \\ \implies \sf no. \: of \: moles = \frac{1.5}{208.23} \\ \\ \sf \implies no. \: of \: moles = 0.0072](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20no.%20%5C%3A%20of%20%5C%3A%20moles%20%3D%20%20%5Cfrac%7Bweight%20%5C%3A%20of%20%5C%3A%20solute%7D%7Bmolar%20%5C%3A%20mass%20%5C%3A%20of%20%5C%3A%20solute%7D%20%20%5C%5C%20%20%5C%5C%20%5Cimplies%20%5Csf%20no.%20%5C%3A%20of%20%5C%3A%20moles%20%3D%20%20%5Cfrac%7B1.5%7D%7B208.23%7D%20%20%5C%5C%20%20%5C%5C%20%20%5Csf%20%5Cimplies%20%20no.%20%5C%3A%20of%20%5C%3A%20moles%20%3D%200.0072%20)
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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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![\sf \hookrightarrow molality = \frac{no.\: of \: moles}{weight \: of \: solution} \\ \\ \sf \hookrightarrow molality = \frac{0.072}{1.5} \\ \\ \sf \hookrightarrow molality = 0.048 \: mol {kg}^{ - 1}](https://tex.z-dn.net/?f=%20%5Csf%20%20%5Chookrightarrow%20molality%20%3D%20%20%5Cfrac%7Bno.%5C%3A%20of%20%5C%3A%20moles%7D%7Bweight%20%5C%3A%20of%20%5C%3A%20solution%7D%20%20%5C%5C%20%20%5C%5C%20%20%5Csf%20%20%5Chookrightarrow%20molality%20%3D%20%20%5Cfrac%7B0.072%7D%7B1.5%7D%20%20%5C%5C%20%20%5C%5C%20%20%5Csf%20%20%5Chookrightarrow%20molality%20%3D%200.048%20%5C%3A%20mol%20%7Bkg%7D%5E%7B%20-%201%7D%20)
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![\textsf{ \large{ \underline{Now substituting the required values}}}](https://tex.z-dn.net/?f=%20%5Ctextsf%7B%20%5Clarge%7B%20%5Cunderline%7BNow%20substituting%20the%20required%20values%7D%7D%7D%20)
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![\sf \longmapsto \Delta T_b = 3 \times 0.51 \times 0.0048 \\ \\ \\ \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20%5CDelta%20T_b%20%3D%203%20%20%5Ctimes%200.51%20%20%5Ctimes%200.0048%20%5C%5C%20%20%5C%5C%20%5C%5C%20%20%20%20%20%5Cboxed%7B%20%5Ctt%7B%20%5Clongmapsto%20%5CDelta%20T_b%20%3D0.00735%7B%20%5Cdegree%7DC%7D%7D)
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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>
<em>hey, im jordan :)</em>
the SI unit for the mass of subatomic particles is <u>amu (atomic mass unit)</u>
<em>hope this helps!</em>
<em>have a great day :D</em>
The balanced thermochemical equation is
KBr ------- K + 1/2 Br2
<h3>What is thermochemical equation? </h3>
A Thermochemical Equation is defined as the balanced stoichiometric chemical equation which includes the enthalpy change, ΔH.
The chemical equation for the decomposition of potassium bromide to its constituent elements bromine ans potassium :
KBr ----- K + Br2
The balanced thermochemical equation of the decomposition of potassium bromide to its constituent elements potassium and bromide as follows
KBr ------- K + 1/2 Br2
As the heat is absorbed in this reaction therefore, heat is positive.
Thus, we concluded that the balanced thermochemical equation is
KBr ------- K + 1/2 Br2
learn more about thermochemical equation:
brainly.com/question/2733624
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Answer:
C
Explanation:
a variable does that dose not depend on that of another