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pshichka [43]
1 year ago
13

Describe the general properties of rotations. Include a discussion of the properties of rigid transformations, and line segments

connecting corresponding points to each other and to the center of rotation.
Mathematics
2 answers:
Alex777 [14]1 year ago
4 0

Sample Response: Rotations are rigid transformations, which means they preserve the size, length, shape, and angle measures of the figure. However, the orientation is not preserved. Line segments connecting the center of rotation to a point on the pre-image and the corresponding point on the image have equal length. The line segments connecting corresponding vertices are not parallel.


Yanka [14]1 year ago
4 0

Answer:

Sample Response: Rotations are rigid transformations, which means they preserve the size, length, shape, and angle measures of the figure. However, the orientation is not preserved. Line segments connecting the center of rotation to a point on the pre-image and the corresponding point on the image have equal length. The line segments connecting corresponding vertices are not parallel.

Step-by-step explanation:

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What is the logarithm of the equilibrium constant, log K, at 25°C of the voltaic cell constructed from the following two half-re
sashaice [31]

Answer:

4.0921 is the logarithm of the equilibrium constant.

Step-by-step explanation:

Fe^{2+} (aq) +2e^{-}\rightarrow Fe(s); ​E​° = - 0.41 V

Ag^+(aq) + e^-\rightarrow Ag(s); E° = 0.80 V

Iron having negative value of reduction potential .So ,that means that it will loose electron easily and get oxidized.Hence, will be at anode.

E^{o}_{cell}=Reduction potential of cathode - Reduction potential of anode

E^{o}_{cell}=E^{o}_c-E^{o}_a

=0.80 V-(-0.41 V)=1.21 V

Fe^{2+} (aq) + 2e^{-}\rightarrow Fe(s); ​E​° = - 0.41 V

2Ag^+(aq) + 2e^-\rightarrow 2Ag(s); E° = 0.80 V

Net reaction: Fe(s)+2Ag^{+}\rightarrow Fe^{2+}+2Ag(s)

n = 2

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

\Delta G^o=-nfE^o_{cell}

and,

\Delta G^o=-RT\ln K_{eq}

Equating these two equations, we get:

nfE^o_{cell}=RT\ln K_{eq}

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

E^o_{cell} = standard electrode potential of the cell = 1.21 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = 25^oC=[273+25]=298K

Putting values in above equation, we get:

2\times 96500\times 1.21 V=8.314\times 298\times \ln K_{eq}

\ln K_{eq}=9.3478

\log K_{eq}=\frac{9.3478}{2.303}=4.0921

4.0921 is the logarithm of the equilibrium constant.

7 0
1 year ago
In complete sentences, explain whether or not it makes sense to say that the density of helium is 0.1785 kilograms per square me
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2 years ago
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Answer:

x=-3

Step-by-step explanation:

(3x-15)/2 = 4x

Multiply each side by 2

(3x-15)/2 *2= 4x*2

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Subtract 3x from each side

3x-15-3x = 8x-3x

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Divide each side by 5

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5 0
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8 0
1 year ago
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