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3 years ago

8

The reaction described in Part A required 3.54 L of magnesium chloride. What is the concentration of this magnesium chloride sol

ution?

1 answer:

labwork [276]3 years ago

7 0

we need the cholride solution Part B to answer your question

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1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.

Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice ) = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

$\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}$

$\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}$

$\Delta \ S = {(1*37.7)In \dfrac{263}{273}$

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0

3 years ago

Which process releases the energy that eventually produces lighting in a thunderstorm

Elena L [17]

The process that releases energy which causes lightning is known as Glaciation

5 0

3 years ago

Which of these elements has a smaller atomic #? 1=aluminum 2=beryllium 3=tungsten

enot [183]

2
refer to periodic table for explanation

6 0

3 years ago

Initially, a particular sample has a total mass of 200 grams and contains 128 x 1010 radioactive nuclei. These radioactive nucle

lubasha [3.4K]

Explanation:

Formula to calculate how many particles are left is as follows.

N = $N_{0} (\frac{1}{2})^{l}$

where, $N_{0}$ = number of initial particles

l = number of half lives

As it is given that number of initial particles is $128 \times 10^{10}$ and number of half-lives is 3.

Hence, putting the given values into the above equation as follows.

N = $N_{0} (\frac{1}{2})^{l}$

= $128 \times 10^{10}(\frac{1}{2})^{3}$

= $16 \times 10^{10}$

or, $1.6 \times 10^{11}$

Thus, we can conclude that $1.6 \times 10^{11}$ particles of radioactive nuclei remain in the given sample.

In five hours we've gone through 5 half lives so the answer is:

particles

8 0

3 years ago

Balance the following Equation:

pashok25 [27]

Answer:

HCl

Explanation:

Given data:

Mass of Zn = 50 g

Mass of HCl = 50 g

Limiting reactant = ?

Solution:

Chemical equation:

Zn + 2HCl → ZnCl₂ + H₂

Number of moles of Zn:

Number of moles = mass / molar mass

Number of moles = 50 g/ 65.38 g/mol

Number of moles = 0.76 mol

Number of moles of HCl:

Number of moles = mass / molar mass

Number of moles = 50 g/ 36.5 g/mol

Number of moles = 1.4 mol

Now we will compare the moles of Reactant with product.

Zn : ZnCl₂

1 : 1

0.76 : 0.76

Zn : H₂

1 : 1

0.76 : 0.76

HCl : ZnCl₂

2 : 1

1.4 : 1/2×1.4 = 0.7

HCl : H₂

2 : 1

1.4 : 1/2×1.4 = 0.7

Less number of moles of product are formed by HCl it will act limiting reactant.

6 0

3 years ago