All categories

2 years ago

The melting point of water is 0°C at 1 atm pressure because under these conditions:

Chemistry

1 answer:

Tems11 [23]2 years ago

8 0

Answer:

The correct answer is option C, that is, ΔS and ΔSsurr for the process H2O (s) ⇒ H2O(l) are equal in magnitude and opposite in sign.

Explanation:

The temperature at which solid state of water get transformed into liquid state is termed as the melting point of 0 °C. It can be shown by the reaction:

H2O (s) ⇒ H2O (l)

The degree of randomness of a molecule is known as entropy. With the transformation of ice into liquid state, there is an increase in randomness. Thus, the value of entropy becomes positive as shown:

Entropy change (ΔSsys) = ΔSproduct - ΔSreactant

= (69.9 - 47.89) J mol/K

= 22.0 J mol/K

Therefore, the value of entropy change is positive.

Now the value of entropy for surrounding ΔSsurr will be,

ΔSsurr = -ΔHfusion/T

= -6012 j/mol/273

= -22.0 J/molK

Hence, the value of ΔSsurr and ΔSsys exhibit same magnitude with opposite sign.

You might be interested in

A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit

Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

$\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}$

$P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}$

$P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times 35.0} = 0.776969038566$

∴ The final pressure P₂ ≈ 0.78 atm.

4 0

2 years ago

When 1.00 g of boron is burned in o2(g) to form b2o3(s), enough heat is generated to raise the temperature of 733 g of water fro

Bas_tet [7]

<span>Answer: For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees. 4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ. Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work. To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3. .0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.</span>

7 0

2 years ago

One of the emission spectral lines for Be31 has a wavelength of 253.4 nm for an electronic transition that begins in the state w

max2010maxim [7]

Explanation:

The given data is as follows.

$\lambda$ = 253.4 nm = $253.4 \times 10^{-9}m$ (as 1 nm = $10^{-9}$ )

$n_{1}$ = 5, $n_{2}$ = ?

Relation between energy and wavelength is as follows.

E = $\frac{hc}{\lambda}$

= $\frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{253.4 \times 10^{-9}}$

= $0.0784 \times 10^{-17}$ J

= $7.84 \times 10^{-19} J$

Hence, energy released is $7.84 \times 10^{-19} J$ .

Also, we known that change in energy will be as follows.

$\Delta E = -2.178 \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}$

where, Z = atomic number of the given element

$7.84 \times 10^{-19} J = -2.178 \times (4)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}$

$\frac{7.84 \times 10^{-19} J}{34.848} = \frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}$

0.02 + 0.04 = $\frac{1}{n^{2}_{1}}$

$n_{1}$ = $\sqrt{\frac{1}{0.06}}$

= 4

Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.

4 0

2 years ago

Monochlorination of propane yields two constitutional isomers, and dichlorination yields four. Trichlorination yields five const

GaryK [48]

By Tri chlorination, three chlorine atoms can replace the 3 hydrogen atoms in propane molecule to give tri-chlorinated products according to the attached picture.