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larisa
2 months ago
9

If LK=MK, LK=7x-10,KN=x+3, MN=9x-11 and KJ=28 find LJ

Mathematics
1 answer:
zzz [12.3K]2 months ago
3 0
Consider points J, K, L, M, and N located on a single straight line.

Calculate MK by subtracting KN from MN: MK = MN + (-KN) = MN - KN = 9x - 11 - (x + 3) = 8x - 14

Given that LK equals MK and LK is also 7x - 10, we set:

7x - 10 = 8x - 14

Subtracting 7x from both sides gives:

0 - 10 = x - 14

Adding 10 to both sides results in:

0 = x - 4

Hence, x equals 4.

Next, compute LJ as the sum of MK and KJ.

Since MK = LK = 7x - 10, plugging in x = 4 yields MK = 7(4) - 10 = 28 - 10 = 18

Therefore, LJ equals 18 plus 28, totaling 46.
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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
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

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

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
What algebraic expression represents GK, if GK =37, what are GH and JK. GJ=4x+2 and HK=6x-1
Inessa [12570]
The outcome is "GK=37, GH=14, and JK=19".

The problem references a missing attachment file, thus please find it attached.

The provided values are:

GJ=4x+2

HK=6x-1

Given GK =37

Need to calculate:

HJ=?

From GK: GK = GJ+JK yields GJ+(HK-HJ)

= 4x+2+((6x-1)-x) yields

9x+1...(a)Substituting (a) into the equation gives: 9(4)+1 yields 37 solves to x=4... for answer (b) and (c). Substituting x=4 yields:

GH=14 and JK=19.
8 0
1 month ago
A 5 1/2 quart pot is filled 2/3 of the way with water. How many more quarts of water can the pot hold?
Inessa [12570]

The pot has the capability to hold 1/3 more than its current level. Therefore, 1/3 of 5 1/2 quarts equals:

(1/3)(11/2) = 11/6 quarts. Thus, the total capacity of the pot is 11/6 quarts.

8 0
2 months ago
Read 2 more answers
The parent cosecant function is shifted 4 units right and 3 units up. Which of the following is the graph of the transformed fun
AnnZ [12381]
We can use a graphing tool to plot the <span> cosecant function
</span>check the attached image

the answer corresponds to option B

4 0
1 month ago
Read 2 more answers
westfalls is 7 miles south of edenville and concord is 13 miles west of westfalls. what is the distance from edenville to concor
Leona [12618]

Response:

The distance from Endnville to Concord is calculated as 14.8 miles

Step-by-step explanation:

∵ Westfalls is located 7 miles south of Endnville

∵ Concord is situated 13 miles west of Westfalls

∵ South and West are ⊥

∴ The total distance from Endnville to Concord is √(7)² + (13)²

                                                                      = 14.8 miles

7 0
1 month ago
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