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saveliy_v [14]
3 years ago
11

Which atom has five electrons in its outer level and 10 electrons in the inner energy levels

Chemistry
1 answer:
Sever21 [200]3 years ago
6 0

Oxygen 47 Hydrogen 63

Silicon 28 Oxygen 25.5

Aluminum 7.9 Carbon 9.5

Iron 4.5 Nitrogen 1.4

Calcium 3.5 Calcium 0.31

Sodium 2.5 Phosphorus 0.22

Potassium 2.5 Chlorine 0.03

Magnesium 2.2 Potassium 0.06

Titanium 0.46 Sulfur 0.05

Hydrogen 0.22 Sodium 0.03

Carbon 0.19 Magnesium 0.01

All others <0.1 All others <0.01 Living matter

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     Using the Fundamental Equation of Calorimetry, we have:

Q=mc\Delta T \\ Q=59.7.0.231.(100-25) \\ \boxed {Q=1034.3025J}      

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7 0
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Which of the following best describes the particles in an atom?
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Answer:

Electrons move around a nucleus.

Explanation:

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When considering light moving through a diffraction grating, you should treat the light as what?
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Answer:

A particle

Explanation:

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5 0
3 years ago
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If you needed to make 100 mL of a 0.2 M fruit drink solution from the 1.0 M fruit drink solution, how would you do it? (Hint: Us
grigory [225]

Answer:

We take 20.0 mL of the 1.0 M fruit drink solution and then add 80.0 mL of water to  make 100 mL of a 0.2 M fruit drink solution.

Explanation:

  • Using the rule that: the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution.

<em>(MV) before dilution = (MV) after dilution.</em>

M before dilution = 1.0 M, V before dilution = ??? mL.

M after dilution = 0.2 M, V after dilution = 100 mL.

<em>∴ V before dilution = (MV) after dilution / M before dilution </em>= (0.2 M)(100 mL) / (1.0 M) = <em>20.0 mL.</em>

<em>So, we take 20.0 mL of the 1.0 M fruit drink solution and then add 80.0 mL of water to  make 100 mL of a 0.2 M fruit drink solution.</em>

8 0
3 years ago
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Determine the equilibrium constant, Keq, at 25°C for the reaction
adelina 88 [10]

Explanation:

The given chemical reaction is:

2Br^- (aq) + I_2(s)  Br_2(l) + 2I^- (aq)

E^ocell=oxidation potential of anode + reduction potential of cathode\\

The relation between Eo cell and Keq is shown below:

deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell

The value of Eo cell is:

Br- undergoes oxidation and I2 undergoes reduction.

Reduction takes place at cathode.

Oxidation takes place at anode.

Hence,

E^ocell= (-1.07+0.53)V\\=-0.54V

F=96485 C/mol

n=2 mol

R=8.314 J.K-1.mol-1

T=298K

Substitute all these values in the above formula:

ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3

Answer:

Keq=6.13x10^33

3 0
3 years ago
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