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3 years ago

Mixtures are 'separated through ._changes. compound physical mixtures

Chemistry

1 answer:

Neporo4naja [7]3 years ago

8 0

Answer:

physical changes

Explanation:

mixtures are the combination of two or more components through physical means and can be broken down or reversed into its initial state

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A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both sol

Reika [66]

Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

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

$q=[q_1+q_2]$

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

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the solution

$c_1$ = specific heat of calorimeter = $12.1J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_2$ = mass of water or solution = $Density\times Volume=1/mL\times 100.0mL=100.0g$

$\Delta T$ = change in temperature = $T_2-T_1=(26.3-20.2)^oC=6.1^oC$

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

$q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]$

$q=2623.61J$

Now we have to calculate the enthalpy change for the process.

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

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 2626.61 J

n = number of moles of copper sulfate used = $Concentration\times Volume=1M\times 0.050L=0.050mole$

$\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole$

Therefore, the enthalpy change for the process is 52.5 kJ/mole.

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3 years ago

A) NaCl2 b) Al3Br c) AlO2 d) H2Br e) CaCl2

Elena L [17]

The answer to this question is C) ALO2

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3 years ago

The free energy of formation of nitric oxide, NO, at 1000 K (roughly the temperature in an automobile engine during ignition) is

Assoli18 [71]

Answer: The value of $K_p$ for the chemical equation is $8.341\times 10^{-5}$

Explanation:

For the given chemical equation:

$N_2(g)+O_2(g)\rightarrow 2NO(g)$

To calculate the $K_p$ for given value of Gibbs free energy, we use the relation:

$\Delta G=-RT\ln K_p$

where,

$\Delta G$ = Gibbs free energy = 78 kJ/mol = 78000 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = temperature = 1000 K

$K_p$ = equilibrium constant in terms of partial pressure = ?

Putting values in above equation, we get:

$78000J/mol=-(8.314J/Kmol)\times 1000K\times \ln K_p\\\\Kp=8.341\times 10^{-5}$

Hence, the value of $K_p$ for the chemical equation is $8.341\times 10^{-5}$

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4 years ago

At 25 degrees, the rate constant for the first order decomposition of a pesticide solution is 6.40 x 10^-2 min^-2. If the starti

EastWind [94]

If the initial amount is 1, the amount remaining at time t is exp(-rt), where r is the rate constant (.0064), and t is the time (62). The value of the exponential is 0.6725, and doing the obvious multiplication gives .0211 mole.

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3 years ago

How to find the valence electrons on a periodic table + energy levels?

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