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Natasha_Volkova [10]
2 years ago
10

What example of mixture could be seperated using two or more techniques? Explain your answer

Chemistry
1 answer:
STatiana [176]2 years ago
8 0
You can take two liquids of different densities (how much mass is in a given volume) and pour them into a funnel. An example is oil and water. When the mixture settles, the denser liquid will be at the bottom, and drips through the funnel first. This is a separation that you can just let occur naturally.
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Which tool was most likely used in a procedure if the lab report shows that approximately 300 mL of water was used?
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When light rays appear to be coming from behind a mirror, what type of image has been formed?
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Read 2 more answers
If 25.8 grams of BaO dissolve in enough water to make a 212-gram solution, what is the percent by mass of the solution?
Igoryamba
The correct answer is 12.2% BaO.

The solution is found by dividing the mass of the BaO, which is 25.8 grams, by the total mass of the solution, which is 212 grams, then multiplying it by 100 to get the percentage:
\frac{25.8}{212}*(100) =12.2%
3 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
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