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In the laboratory a general chemistry student finds that when 2.84 g of KClO4(s) are dissolved in 107.70 g of water, the tempera

vredina [299]

Answer : The enthalpy change of dissolution of $KClO_4$ is 54.3 kJ/mole

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $1.55J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 107.70 g

$\Delta T$ = change in temperature = $T_2-T_1=(22.80-20.34)=2.46^oC$

Now put all the given values in the above formula, we get:

$q=[(1.55J/^oC\times 2.46^oC)+(107.70g\times 4.184J/g^oC\times 2.46^oC)]$

$q=1112.3J=1.1123kJ$

Now we have to calculate the enthalpy change of dissolution of $KClO_4$

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 1.1123 kJ

m = mass of $KClO_4$ = 2.84 g

Molar mass of $KClO_4$ = 138.55 g/mol

$\text{Moles of }KClO_4=\frac{\text{Mass of }KClO_4}{\text{Molar mass of }KClO_4}=\frac{2.84g}{138.55g/mole}=0.0205mole$

$\Delta H=\frac{1.1123kJ}{0.0205mole}=54.3kJ/mole$

Therefore, the enthalpy change of dissolution of $KClO_4$ is 54.3 kJ/mole

5 0

4 years ago

Identify each of the following:

zaharov [31]

Most metallic: Ge

Largest atomic radius: Bi

Highest ionization energy: Cl

Could you mark me as brainliest?

7 0

3 years ago

A sample of an unknown gas takes 222 s to diffuse through a porous plug at a given temperature. At the same temperature, N2(g) t

gulaghasi [49]

Answer:

$\large \boxed{\text{45.1 g/mol}}$

Explanation:

Graham’s Law applies to the diffusion of gases:

The rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

$r \propto \dfrac{1}{\sqrt{M}}$

If you have two gases, the ratio of their rates of diffusion is

$\dfrac{r_{2}}{r_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}$

The time for diffusion is inversely proportional to the rate.

$\dfrac{t_{2}}{t_{1}} = \sqrt{\dfrac{M_{2}}{M_{1}}}$

Data:

t₂ = 222 s

t₁ = 175 s

M₁ = 28.01

Calculation :

$\begin{array}{rcl}\dfrac{222}{175} & = & \sqrt{\dfrac{M_{2}}{28.01}}\\\\1.269 & = & \sqrt{\dfrac{M_{2}}{28.01}}\\\\1.609 & = & \dfrac{M_{2}}{28.01}\\\\M_{2} & = & 1.609 \times 28.01\\ & = & \textbf{45.1 g/mol}\\\end{array}\\\text{The molar mass of the unknown gas is $\large \boxed{\textbf{45.1 g/mol}}$}$

3 0

3 years ago

Which of the following answers below is the correctly balanced half-reaction for the reduction of the SO4 ^-2 ion in acid soluti

Rom4ik [11]

Answer:

C

Explanation:

The oxidation number of Sulphur in SO4^2- is;

x + 4(-2) = -2

x - 8 = -2

x = -2 + 8

x = 6

Now,

the oxidation number of sulphur in H2SO3 is

2 (1) + x + 3(-2) = 0

2 + x -6 = 0

-4 + x = 0

x = 4

Hence, the oxidation number of sulphur changed from +6 to +4 which signifies gain of two electrons as shown in option C.

8 0

3 years ago

Scientists publish a revised list of dangerous greenhouse gases. Which statements describe the importance of this information? C

Pie

Answer:

It leads to new jobs that search for innovative ways to control emissions.

Explanation:

Publishing a report that is a revised list of green house gases by scientists underscores the importance of new new jobs that seek ways of controlling emissions. The publication will inspire scientific research into control of emission of greenhouse gases. This will create new jobs for scientists who will be charged with the responsibility of ensuring that emission of Tue newly discovered greenhouse gasses is greatly reduced.

Several scientists earn a living by actively carrying out research on control of the emission of greenhouse gasses.

8 0

4 years ago