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:

Explanation:
Given that:
Half life = 30 min
Where, k is rate constant
So,
The rate constant, k = 0.0231 min⁻¹
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t
is the initial concentration
Given that:
The rate constant, k = 0.0231 min⁻¹
Initial concentration
= 7.50 mg
Final concentration
= 0.25 mg
Time = ?
Applying in the above equation, we get that:-

Mineral composition affects the classification of igneous rock. in simplified classification, igneous to is are classified by the type of feldspar present, by the I type a of feldspar present, or the absence of quartz. in case of neither present, then by the type of iron or magnesium present.
Answer:
0.12M
Explanation:
A balanced equation for the reaction will go a great deal in obtaining our desired result. So, let us write a balanced equation for the reaction
HCl + NaOH —> NaCl + H2O
From the above equation,
nA (mole of the acid) = 1
nB (mole of the base) = 1
Data obtained from the question include:
Vb (volume of the base) = 30mL
Mb (Molarity of the base) = 0.1M
Va (volume of the acid) = 25mL
Ma (Molarity of the acid) =?
The molarity of the acid can be obtained as follow:
MaVa/MbVb = nA/nB
Ma x 25/ 0.1 x 30 = 1
Cross multiply to express in linear form
Ma x 25 = 0.1 x 30
Divide both side by 25
Ma = (0.1 x 30) / 25
Ma = 0.12M
The molarity of the acid is 0.12M
Answer:
a) 
b) entropy of the sistem equal to a), entropy of the universe grater than a).
Explanation:
a) The change of entropy for a reversible process:


The energy balance:
![\delta U=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=%5Cdelta%20U%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
If the process is isothermical the U doesn't change:
![0=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=0%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)


The work:

If it is an ideal gas:


Solving:

Replacing:


Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.