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allochka39001 [22]
3 years ago
5

31. A mixture with sand, pink sugar and water would be best separated using

Chemistry
1 answer:
ozzi3 years ago
3 0

Answer:

Filtration and evaporation

Explanation:

A mixture of sand, pink sugar and water can be separated by filtration and evaporation.

The pink sugar is already dissolved in the water but the sand remains undissolved in the water hence it can be filtered off. The filtrate now contains the sugar.

The sugar is obtained from the filtrate by evaporation of the solution to yield the solid pink sugar and steam which can be condensed again to give liquid water.

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B. The sand increases friction by increasing roughness.
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pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles of a diatomic gas (which beha
Fiesta28 [93]

Complete Question

Questions Diagram is attached below

Answer:

*  W=1142.86Joule

*  Q=997.7J

*  H=2140.5J

Explanation:

From the question we are told that:

Temperature T=337K

Pressure P=(60-55)Pa*10^5

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\frac{P_2}{P_1}=\frac{V_1}{V_2}^n

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Therefore

Work-done

 W=\int{pdv}

 W=\frac{55*10^5*1.6*10^{-3}*60*10^5*1.4*10^{-3}}{1-0.65}

 W=1142.86Joule

Generally the equation for internal energy is mathematically given by

 Q=mC_vdT\\\\Q=\frac{3*1*3.314*16}{1.4-1}

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2 years ago
His transmitter is mostly located in diffuse neuronal systems in the CNS, with cell bodies particularly in the raphe nuclei. It
joja [24]

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3 years ago
Iron(II) sulfide has a primitive cubic unit cell with sulfide ions at the lattice points.
masya89 [10]

We have that the  the density of FeS  is mathematically given as

  •  \phi=2.56h/cm^3

From the question we are told

Iron(II) sulfide has a primitive <em>cubic</em> unit cell with <em>sulfide</em> ions at the <em>lattice points.</em>

The ionic radii of iron(II) ions and sulfide ions are 88 pm and 184 pm, respectively.

What is the density of FeS (in g/cm3)?

<h3>Density</h3>

Generally the equation for the Velocity  is mathematically given as

V_c=a^3=(2\pi Fe^{2+}+2\pi S^{2-})^3\\\\Therefore\\\\V=a^3(2\pi*0.088+2\pi 0.184)^3\\\\V=16.98*10^{-23}\\\\Therefore\\\\\phi=n\frac{PFeion+PSion}{VNa}\\\\\phi=3*\frac{55.85+32}{16.9*10^{-23}*6.023*10^{23}}

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  • \phi=2.56h/cm^3

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3 years ago
What two parts are needed to make a neutral atom of neon
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Answer:

it needs two electrons in the first and eight to fill the second.

Explanation:

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