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stiks02 [169]
3 years ago
6

Will mark brainliest now

Chemistry
1 answer:
Aleks [24]3 years ago
4 0
I think it’s water evaporation I’m not sure tho
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A gas balloon has a volume of 80.0 mL at 300K, and a pressure of 50.0 kPa. If the pressure changes to 80 kPa and the temperature
fomenos

Answer:  The new volume is 53.3 ml

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas =  50.0 kPa

P_2 = final pressure of gas = 80.0 kPa

V_1 = initial volume of gas = 80.0 ml

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 300K

T_2 = final temperature of gas = 320K

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

\frac{50.0\times 80.0}{300}=\frac{80.0\times V_2}{320}

V_2=53.3ml

The new volume is 53.3 ml

4 0
2 years ago
The estimated heat of vaporization of diethyl ether using the Chen's rule is A. 29.7 KJ/mol B. 33.5 KJ/mol C. 26.4 KJ/mol D. 36.
Brums [2.3K]

Answer:

C. 26.4 kJ/mol

Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]

Where,

\Delta H_v is the Heat of vaoprization (J/mol)

T_b is the normal boiling point of the gas (K)

T_c is the Critical temperature of the gas (K)

P_c is the Critical pressure of the gas (bar)

R is the gas constant (8.314 J/Kmol)

For diethyl ether:

T_b=307.4\ K

T_c=466.7\ K

P_c=36.4\ bar

Applying the above equation to find heat of vaporization as:

\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]

\Delta H_v=26400 J/mol

The conversion of J into kJ is shown below:

1 J = 10⁻³ kJ

Thus,

\Delta H_v=26.4 kJ/mol

<u>Option C is correct</u>

6 0
3 years ago
My swimming pool is rectangular (16 feet by 34 feet) and has a depth of 6 feet. Lets imagine that my pool water is full to the t
Reil [10]

Answer:

Number of moles of photons required = 5.04 × 10⁴ moles

Explanation:

The energy of a photon can be calculated from Planck's equation E = hc/λ

Where h = 6.63 × 10-³⁴ Js and c, the velocity of light = 3.0 × 10⁸ m/s

Energy of one mole of photons = N₀ × hc/λ

wavelength of photon, λ = 520 nm = 5.20 × 10-⁷ m

Energy of one mole of photons = 6.02 × 10²³ × 6.63 × 10−³⁴ × 3 × 10⁸/5.20 × 10-⁷

Energy of one mole of photons = 2.30 × 10⁵ J/mol

Energy required to raise the temperature of a given mass of a substance, E = mcΔT

Where m is mass of substance,  c is specific heat capacity,  ΔT is temperature difference

Mass ofnwternin the pool = volume × density

Volume of water = Volume of swimming pool

Volume of water = 16 × 34 × 6 ft³ = 3264 ft³

1 ft³ = 28316.8 cm³; 3264 ft³ = 28316.8 × 3264 = 92426035.2 cm³

Density of water = 1 g/cm³

Mass of water = 92426035.2 cm³ × 1 g/cm³ = 92426035.2g

ΔT = 80°C - 50°C = 30°C, c = 4.18 J/g/K

Energy required to raise 92426035.2 g water by 30° C = 92426035.2 × 4.18 × 30

Energy required = 1.16 × 10¹⁰ J

Hence, number of moles of photons required = 1.16 × 10¹⁰ J/2.30 × 10⁵ J/mol

Number of moles of photons required = 5.04 × 10⁴ moles

5 0
3 years ago
How do you write balance equations Fe2 S3+O2-Fe2O3+SO2
lana66690 [7]

Answer:

2Fe2 S3+6O2 → 2Fe2O3+3SO2

Explanation:

4 0
2 years ago
Consider the formation of the three solutionsshown in the table.
bazaltina [42]

Answer:

sorry please can you snap the diagram ...the question is not clear to my understanding

8 0
2 years ago
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