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2 years ago

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A sample that weighs 107.75 g is a mixture of 30% helium atoms and 70% krypton atoms. How many particles are present in the samp

le?

1 answer:

r-ruslan [8.4K]2 years ago

5 0

<span>A sample that weighs 107.75 g is a mixture of 30% helium atoms and 70% krypton atoms. How many particles are present in the sample? </span>3.59

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Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

ira [324]

Explanation:

The given data is as follows.

m = 10.0 kg = 10,000 g (as 1 kg = 1000 g)

Initial temp. of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Initial temp. of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

Convert temperature into kelvin as (50 + 273) K = 323 K.

Also, we know that the relation between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

$\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= -554.12 J/K + 647.49 J/K

= 93.37 J/K

Thus, we can conclude that for the given reaction $\Delta H$ is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

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2 years ago

Consider the following balanced equation: Fe + 2Ag+ Fe2+ + 2Ag Which statement is true?

vaieri [72.5K]

I’m pretty sure it’s d

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2 years ago

. A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture

postnew [5]

Answer:

A) Mass flow rate of air = 22.892 kmol/hr

B)percentage by mass of oxygen in the product gas = 22.52%

Explanation:

We are given that the mixture containing 9.0 mole% methane in air flowing.

Thus, we have 0.09 mole of methane(CH4) and the remaining will be the air which is (100% - 9%) = 91% = 0.91

Molar mass of CH4 = 12 + 1(4) = 16 g/mol

We are given the average molecular weight of air = 29 g/mol

Thus;

Average molar mass of air and methane mixture is;

M_avg = (0.09 × 16) + (0.91 × 29)

M_avg = 27.83 g/mol

We are told that air flowing at a rate of 7 × 10² kg/h = 700 kg/h

Thus;

Mass flow rate of CH4 in air mixture = 700kg/h × (0.09CH4)/1 mix × (1/27.83kg/kmol) = 2.264 kmol/hr

Mass flow rate of air in mixture = 2.264kmol/h × 0.91kmol air/0.09kmolCH4 = 22.892 kmol/hr

We are told that the mixture is capable of being ignited if the mole percent of methane is between 5% and 15%.

Thus, for 5% of methane, the air required will be;

2.264kmol/h × 0.95kmol air/0.05kmol CH4 = 43.016 kmol/hr

Now, the dilution air needed will be =

43.016 - 22.892 = 20.124 kmol/hr

Total mass flow rate of mixture =

700kg/hr + (20.124kmol/hr × 29kg/mol) = 1283.596 kg/hr

We are told that air consist of 21 mole% Oxygen (O2).

Molar mass of oxygen = 32

Thus;

Mass fraction of oxygen in the product gas = 43.016kmol/h × (0.21molO2/1mol air) × (32kg oxygen/1kmol oxygen) × (1/1283.596kg/h) = 0.2252

Thus, written in percentage form, we have; 22.52%

So, percentage by mass of oxygen in the product gas = 22.52%

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2 years ago

Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the

Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by $\eta$ .

The given data is as follows.

$\eta$ = 0.82, $T_{1}$ = (21 + 273) K = 294 K

$P_{1}$ = 200 kPa, $P_{2}$ = 1000 kPa

Therefore, calculate the final temperature as follows.

$\eta = \frac{T_{2} - T_{1}}{T_{2}}$

0.82 = $\frac{T_{2} - 294 K}{T_{2}}$

$T_{2}$ = 1633 K

Final temperature in degree celsius = $(1633 - 273)^{o}C$

= $1360^{o}C$

Now, we will calculate the entropy as follows.

$\Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}$

For 1 mole, $\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

It is known that for $NH_{3}$ the value of $C_{v}$ = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

$\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

= $0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}$

= 0.0346 kJ/mol

or, = 34.6 J/mol (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

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2 years ago

Stephanie has been doing research on how petroleum is formed. She says that oxygen must be present while the tiny plants and ani

madreJ [45]

Answer:

No, Stephanie is incorrect. Formation of petroleum cannot take place under the presence of oxygen.

Explanation:

Since, the petroleum is fossil product. Fossil fuel are formed under high pressure and temperature with absence of oxygen for longer period. so the way she is performing is completely incorrect. With the presence of oxygen in no way petroleum will be formed. The temperature and pressure should be in different combination for the formation of the petroleum. Along with the layers of sediments to maintain the pressure is required.

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2 years ago

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