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3 years ago

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15.3 g of nano3 were dissolved in 100g of water in a calorimeter. The temperature of the water drops from 25.00°c to 21.56°c. Ca

lculate delta H for the solution process. (kJ/mol)

1 answer:

Arturiano [62]3 years ago

3 0

Answer:

0.259 kJ/mol ≅ 0.26 kJ/mol.

Explanation:

To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat absorbed by ice (Q = ??? J).

m is the mass of the ice (m = 100.0 g).

c is the specific heat of water (c of ice = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 21.56°C - 25.0°C = -3.44°C).

<em>∵ Q = m.c.ΔT</em>

∴ Q = (100.0 g)(4.186 J/g.°C)(-3.44°C) = -1440 J = -1.44 kJ.

<em>∵ ΔH = Q/n</em>

n = mass/molar mass = (100.0 g)/(18.0 g/mol) = 5.556 mol.

∴ ΔH = (-1.44 kJ)/(5.556 mol) = 0.259 kJ/mol ≅ 0.26 kJ/mol.

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Answer: Volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration.

Explanation:

According to the dilution law,

$C_1V_1=C_2V_2$

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$C_2$ = molarity of diluted solution = 0.075 M (1mM=0.001M)

$V_2$ = volume of diluted solution = 10 ml

Putting in the values we get:

$1M\times V_1=0.075M\times 10ml$

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Answer:

The Lewis structures are in image attached.

Explanation:

Lewis symbol is a representation of an element symbol along with its valence electrons around it in the form of dot(s).

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$Cl=1s^22s^22p^63s^23p^5$

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Sodium's atomic number is 11 in which only 1 electrons are present in its valence shell .So in order to gain noble gas stability it will loose 1 electron to completes its octet. In the Lewis symbol no dot shown as sodium has lost its 1 electron.

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In the Lewis symbol no dot shown as magnesium has lost its 2 electrons.

$Mg=1s^22s^23p^63s^2$

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Calcium's atomic number is 20 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

$Ca= 1s^22s^22p^63s^23p^64s^2$

$Ca^{2+}=1s^22s^23p^6^23p^64s^0$

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Potassium's atomic number is 19 in which only 1 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 1 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 1 electron.

$K= 1s^22s^22p^63s^23p^64s^1$

$K^{+}=1s^22s^23p^6^23p^64s^0$

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Bromine's atomic number is 35 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

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Strontium's atomic number is 38 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

$Sr=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2$

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