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3 years ago

How is the way scientist use the word theory different from the way people use it in there everday language

Chemistry

1 answer:

weqwewe [10]3 years ago

8 0

Answer:

Because Scientists Uses hypothesis and try tests and try proving their points while regular normal people use it to just state a statement

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A large flask is evacuated and weighed, filled with argon gas, and then reweighed. When reweighed, the flask is found to have ga

Elden [556K]

Answer:

100.52

Explanation:

from the ideal gas equation PV=nRT

for a given container filled with any ideal gas P and V remains constant.So T is also constant.R is as such a constant.

So n i.e no of moles will also be constant.

no of moles of Ar=3.224/40=0.0806

no of moles of unknown gas=0.0806

molecular wt of unknown gas=8.102/0.0806=100.52

8 0

3 years ago

Witch element has the smallest radius

natali 33 [55]

Answer: Helium has the smallest radius.

Explanation: Helium has the smallest atomic radius. Due to nuclear charges and stuff ect-

8 0

1 year ago

Which of the following is not an example of a science? a.geology b.astrology c.astronomy d.ecology

Rasek [7]

I think it is b.astrology

6 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

The larger the hole in the ozone layer gets,

seraphim [82]

The more UV rays reach the earth

4 0

3 years ago