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IrinaK
26 days ago
14

Given A is between Y and Z and YA=22, AZ=16x, and YZ=166, find AZ

Mathematics
2 answers:
AnnZ [12.3K]26 days ago
8 0


YA + AZ = YZ

22 + 16x = 166

16x = 166 - 22

16x = 144

x = 144/16

x = 9

AZ = 144

babunello [11.8K]26 days ago
4 0

Answer:

AZ = 144 unit              

Step-by-step explanation:

Because A lies between Y and Z, YZ is the straight segment and A is a point on it between Y and Z.

Given YA = 22

AZ = 16x

YZ = 166

Since A is between Y and Z

⇒ YZ = YA + AZ

⇒ 166 = 22 + 16x

⇒ 144 = 16x

⇒ x = 9

Hence, AZ = 16x = 16 × 9 = 144 unit

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A recipe calls for 400 grams of flour. if leena makes one quarter of the recipe, how many kilograms of flour will she need?
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Shane and Abha earned a team badge that required their team to collect no less than 20002000 cans for recycling. Abha collected
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28 days ago
What is the logarithm of the equilibrium constant, log K, at 25°C of the voltaic cell constructed from the following two half-re
lawyer [12517]

Answer:

4.0921 reflects 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 has a negative reduction potential, indicating its tendency to lose electrons and undergo oxidation, and thus it will be at the 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

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

n = 2

To determine the equilibrium constant, we utilize the correlation with Gibbs free energy, as follows:

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

and,

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

Aligning these two equations yields:

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

where,

n = 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 = reaction temperature = 25^oC=[273+25]=298K

Substituting values into the equation, we arrive at:

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 represents the logarithm of the equilibrium constant.

7 0
1 month ago
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