Its magnifying power is: 4X 5X 9X 20X. A 4-inch, f/5 telescope has a 1-inch eyepiece focal. Its magnifying power is 9x. This answer has been confirmed as correct and helpful.
The change in temperature had the greatest effect at changing the volume of the balloon.
<h3>What are the gas laws?</h3>
The gas laws are used to describe the parameters that has to do with gases.
Given that;
P1 = 98.5 kPa
T1 = 18oC or 291 K
V1 = 74.0 dm3
P2 = 7.0 kPa
V2 = ?
T2 = 18oC or 291 K
P1V1/T1 = P2V2/T2
P1V1T2 =P2V2T1
V2= P1V1T2/P2T1
V2 = 98.5 kPa * 74.0 dm3 * 291 K/ 7.0 kPa * 291 K
V2 = 1041.3 dm3
When;
V1 = 1041.3 dm3
T1 = 291 K
V2 = ?
T2 = 80oC or 353 K
V1/T1 = V2/T2
V1T2 = V2T1
V2 = V1T2/T1
V2 = 1041.3 dm3 * 353 K/291 K
V2 = 1263 dm3
The change in temperature had the greatest effect at changing the volume of the balloon.
Given that
V1 = 100 cm^3
T1 = 273 K
P1 = 1.01 * 10^5 Pa
V2 = ?
P2 = 3.00 x 10^-4 Pa
T2 = -180oC or 255 K
V2= P1V1T2/P2T1
V2 = 1.01 * 10^5 Pa * 100 cm^3 * 255 K / 3.00 x 10^-4 Pa * 273 K
V2 = 3.14 * 10^10 cm^3
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I think it is D I'm not positive
The Keq for the reaction N₂ + 3H2 = 2NH3 if the equilibrium concentrations are Keq = 1.5. The correct option is D.
<h3>What is Keq?</h3>
Keq is the ratio of the concentration of reactant to the concentration of the product.
The balanced equation is
N₂ + 3H₂ = 2NH₃
The equilibrium constant is ![\rm \dfrac{[NH_3]^2}{[N_2]\; [H_2]^3}](https://tex.z-dn.net/?f=%5Crm%20%5Cdfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5C%3B%20%5BH_2%5D%5E3%7D)
The given concentrations of the compounds have been:
Ammonia = 3 M
Nitrogen = 1 M
Hydrogen = 2 M
Thus, the correct option is D. Keq = 1.5.
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I can’t see the picture where’s it at
