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3 years ago

Please answer thank you so much!

Chemistry

2 answers:

Allushta [10]3 years ago

6 0

It is mutualism I really hope this helps

soldier1979 [14.2K]3 years ago

3 0

Hiii it’s mutualism hope this helped :)

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What is one wavelength?

Dmitry_Shevchenko [17]

Answer:

It is the distance between two compressions or two rarefactions.

Explanation:

7 0

3 years ago

An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr

Zina [86]

<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

$C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa$

Rate of flow of ideal gas , n = 4 kmol/hr = $\frac{4\times 1000mol}{3600s}=1.11mol/s$ (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)

We know that:

$\Delta U=0$ (For isothermal process)

So, by applying first law of thermodynamics:

$\Delta U=\Delta q-\Delta W$

$\Delta q=\Delta W$ .......(1)

Now, calculating the work done for isothermal process, we use the equation:

$\Delta W=nRT\ln (\frac{P_1}{P_2})$

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = final pressure = 50 kPa

Putting values in above equation, we get:

$\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW$

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

$\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW$

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

4 0

2 years ago

Help me please i put nine points!

Ahat [919]

Answer:

eons.

It would be D. An eon.

3 0

2 years ago

All of the following conditions of STP are true except A. 101.3 kPa B. 273.15 K C. 22.4 L D. 3.81 kPa

kaheart [24]

3.81 kpa is the condition which is not true at STP

According to IUPAC the standard temperature and pressure that is STP the temperature is 273.15 k or 0 degrees celsius . and the absolute temperature of 101.325 Kpa or 1 atm. In addition at STP the volume of ideal gas is 22.4

8 0

3 years ago

What would the volume of a gas be at 150c if had of volume of 693 ml at 45 c

Vlad1618 [11]

Answer:

Explanation:

T1 = 150°C = (150 + 273.15)K = 423.15K

T2 = 45°C = (45 + 273.15)K = 318K

V1 = 693mL = 693cm³

Applying Charle's law, the volume of a given gas is directly proportional to is temperature provided that pressure remains constant.

V = kT

V1 / T1 = V2 / T2

693 / 423.15 = V2 / 318

V2 = (693 * 318) / 423.15 = 520.79cm³

The new volume of the gas is 520.79cm³

6 0

3 years ago