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irakobra [83]
2 years ago
16

2 horizontal and parallel lines are intersected by 2 diagonal lines to form a triangle with exterior angles. The top angle of th

e triangle is 47 degrees. The exterior angle created by the right side of the triangle and the top horizontal line is x degrees. The bottom left angle of the triangle is y degrees. The exterior angle of the bottom right angle is z degrees above the horizontal line and 63 degrees below. Which statements about the diagram are true? Select three options. x = 63 y = 47 z = 117 x + y = 180 x + z = 180
Chemistry
1 answer:
Sophie [7]2 years ago
5 0

Answer:

x = 63

z = 117

x + z = 180

Explanation:

edg 2020

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When magnesium ribbon is placed in sulfuric acid, a chemical reaction takes place and heat energy is given off. Liang and Naraba
kolezko [41]

Answer:

this is the answer

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6 0
1 year ago
A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.
Sonja [21]

Answer:

Equilibrium constant for PCl_5 is 0.5

Equilibrium constant for decomposition of NO_2 is 1.79 \times 10^{-14}

Explanation:

PCl_5 dissociates as follows:

                    PCl_5 \rightleftharpoons PCl_3+Cl_2

initial          0.72 mol     0         0

at eq.     0.72 - 0.40   0.40      0.40

Expression for the equilibrium constant is as follows:

k=\frac{[PCl_3][Cl_2]}{[PCl_5]}

Substitute the values in the above formula to calculate equilibrium constant as follows:

k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5

Therefore, equilibrium constant for PCl_5 is 0.5

Now calculate the equilibrium constant for decomposition of  NO_2

It is given that 3.3 \times 10^{-3} \% is decomposed.

NO_2 decomposes as follows:

                                  2NO_2 \rightleftharpoons 2NO + O_2

initial                            1.0 M       0           0

at eq. concentration of  NO_2   is:

[NO_2]_{eq}=1-(0.000066) = 0.999934\ M

[NO]_{eq}=6.6 \times 10^{-5}\ M

[O_2]_{eq}=3.3\times 10^{-5} = 3.3\times 10^{-5}\ M      

Expression for equilibrium constant is as follows:

K=\frac{[NO]^2[O_2]}{[NO_2]^2}

Substitute the values in the above expression

K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}

Equilibrium constant for decomposition of NO_2 is 1.79 \times 10^{-14}

8 0
2 years ago
How many ATP are produced from a fatty acid that is 14 carbons long?
fenix001 [56]

Answer:

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Answer:

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