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qaws [65]
3 years ago
6

You want to sweeten a cup of tea, and you have three types of sugar: sugar cubes (volume: 5ml), granulated sugar (volume: 0.1 ml

), and powdered sugar (volume: 0.0001 ml). Which sugar would dissolve the fastest? Why?
Chemistry
1 answer:
sp2606 [1]3 years ago
4 0

Answer:

powdered sugar

Explanation:

The higher is the exposed area of sugar, the faster is the dissolution process. Thus, to choose between the different types of sugar, we have to look at the volume occupied by the sugar.

In sugar cubes, the particles of sugar as compacted in a cube, so the particles inside the cube are not exposed to the solvent (water). So, sugar cubes have the slowest dissolution process. Then, in granulated sugar, the particles have more area exposed, so this type of sugar will dissolve faster than sugar cubes. Finally, powdered sugar is composed of tiny particles with more are exposed, so powdered sugar has the fastest dissolution process.

Therefore, powdered sugar will dissolve the fastest.

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What information is needed to determine the energy of an electron in a many-electron atom? a) n. b) l. c) ml. d)ms
mel-nik [20]

Answer:

The correct answer is b.

Explanation:

The quantum number n specifies the energetic level of the orbital, the first level being the one with the least energy. As n increases, the probability of finding the electron near the nucleus decreases and the orbital energy increases.

In the case of atoms with more than one electron, the quantum number l also determines the sublevel of energy in which an orbital is found, within a certain energy level. The value of l is designated by the letters s, p, d, and f.

Have a nice day!

8 0
3 years ago
Read 2 more answers
Calculate the energy, in joules, required to ionize a hydrogen atom when its electron is initially in the n =2 energy level. The
qaws [65]

Answer:

E_{ionization}=5.45\times 10^{-19}\ J

Explanation:

E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

So, n_i=2 and n_f=\infty (As the hydrogen has to ionize)

Thus,

\Delta E=2.18\times 10^{-18}(\frac{1}{2^2} - \dfrac{1}{{\infty}^2})\ J

\Delta E=2.18\times 10^{-18}(\frac{1}{2^2})\ J

E_{ionization}=5.45\times 10^{-19}\ J

4 0
3 years ago
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

VolumeV=(1.6-1.4)m^3*10^{-3}

Generally the equation for gas Constant is mathematically given by

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

 \frac{55*10^5}{60*10^5}=\frac{1.4*10^{-3}}{1.6*10^{-3}}^n

 n=0.65

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}

 Q=997.7J

Therefore

 H=Q+W

 H=997.7J-11.42.9

 H=2140.5J

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2 years ago
When a radioactive isotope undergoes Beta Decay, which of the following will NOT occur? PLEASE HELP ASAP! THANKS.
sleet_krkn [62]

D) Energy is released


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SEP Engage in Argument A classmate claims that state of matter is an extensive property because it can vary with temperature. Fo
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Answer: I agree with the student because in the question prior to this One question stated thatExtensive properties very with the amount of matter ina sample, so yes i agree.

Explanation:

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2 years ago
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