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2 years ago

8) If you apply a force of 50N to an object, but the object doesn't move,

Chemistry

1 answer:

vlada-n [284]2 years ago

7 0

0 J

Work = Force * Displacement

Since displacement is 0 m (object isn't moving), work would also be 0.

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4. A sample of gas has a pressure of 700 mmHg and 30.0°C. Ar what temperature would the pressure be 600 mmHg if the volume remai

shtirl [24]

Answer:

T₂ = 259.84 K

T₂ = -13.31 °C

Explanation:

Given data:

Initial pressure = 700 mmHg

Initial temperature = 30.0°C (30+273.15 K = 303.15 K)

Final temperature = ?

Final pressure = 600 mmHg

Solution:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

700 mmHg /303.15 K = 600 mmHg / T₂

T₂ = 600 mmHg × 303.15 K / 700 mmHg

T₂ =181890 mmHg.K /700 mmHg

T₂ = 259.84 K

Temperature in celsius

259.84 K - 273.15 = -13.31 °C

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]

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

Explanation:

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

3 years ago

How to describe a ufo??

erastova [34]

Answer:

Unidentified flying object or fly saucer.

Explanation:

6 0

2 years ago

Read 2 more answers

Calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. express the nu

bagirrra123 [75]

To calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. We will simply inspect the "number" of aluminum, sulfur, and oxygen atoms available per one mole of the compound. Here we have Al2(SO4)3, which means that for every mole of aluminum sulfate, there are 2 moles of aluminum, 3 (1 times 3) moles of sulfur, and 12 (4x3) moles of oxygen. Since we have four moles of Al2(SO4)3 given, we simply multiply 4 times the moles present per 1 mole of the compound. So we have 4x2 = 8 moles of Al, 4x3 = 12 moles of sulfur, and 4x12 = 48 moles of oxygen.

So the answer is:
8,12,48

3 0

3 years ago

Read 2 more answers

There are 6 oxygen atoms in a half dozen water (h2o) molecules. how many hydrogen atoms are in the same half dozen molecules?

dlinn [17]

There are 12 hydrogen atoms in 6(H2O)

5 0

3 years ago