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3 years ago

Is it really possible for humans not to alter the natural world around them?

Chemistry

1 answer:

tankabanditka [31]3 years ago

6 0

Answer: I'm sure it's possible but we too lazy

Explanation:

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What is the total number of atoms in 0.20 mol of propanone, CH3COCH3?

Nutka1998 [239]

Answer:

1.2×10²³ atoms.

Explanation:

Data obtained from the question include:

Mole of propanone = 0.20 mole

Number of atoms of propanone =.?

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ atoms.

This implies that 1 mole of propanone also contains 6.022×10²³ atoms.

Thus, we can obtain the number of atoms in 0.20 mole of propanone as illustrated below:

1 mole of propanone contains 6.022×10²³ atoms.

Therefore, 0.20 mole of propanone will contain = 0.2 × 6.022×10²³ = 1.2×10²³ atoms.

Thus, 0.20 mole of propanone contain

1.2×10²³ atoms.

6 0

3 years ago

Some studies have found that when a person mimics an angry expression, their temperature actually rises.

posledela

The correct answer is A) True

5 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

3 years ago

Which of the following is true for balancing equations? A. The number of products should be equal to the number of reactants. B.

AlexFokin [52]

D.) There must be an equal number of atoms of each element on both sides of the equation

5 0

3 years ago

Read 2 more answers

How does active transport differ from passive transport

pickupchik [31]

Active is more active which means it's in person

5 0

3 years ago

Read 2 more answers