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4 years ago

What are two problems with building a refracting telescope with a 10-meter diameter lens?

Chemistry

1 answer:

Lena [83]4 years ago

3 0

Answer:

1. Weight.

2. Large lens will tend to deflect under its own weight

Explanation:

Hello,

Weight is a big part of it. There’s a reason the largest working refractor on Earth (the Clark refractor at Yerkes Observatory in Wisconsin) has a 102-cm objective; a 125-cm lens was created for the Paris Exposition of 1900, but the accompanying telescope was a pain to use (very hard to aim) and was scrapped when no one wanted to buy it.

Lens can only be supported at its edges unlike mirrors, and a large lens will tend to deflect under its own weight unless it’s so thick that it won’t transmit much light. It is also extremely difficult to cast and polish a glass blank of such huge size, which is why (see List of largest optical refracting telescopes - Wikipedia) almost no one has tried building one in over a century.

Best regards.

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Why are pure liquids and solids not included in the reaction quotient or the equilibrium constant expression for a given heterog

AfilCa [17]

Pure liquids and solids at the equilibrium stage:

Pure solids and pure liquids are not included in expressions for the equilibrium constant of heterogeneous equilibria since their concentrations are constant across time. Therefore, including it in our equilibrium expression is not significant.

Equilibrium expression:

When a certain chemical reaction is in equilibrium, the rates of the forward and backward reactions are equal, which means that there is no net change in the concentrations of the reactants and products and that the reaction mixture's composition does not change.

The concentration of solid and liquid:

Solid or liquid concentration,

= No. of Moles/ Volume in L

= Mass/(Volume × molar mass)

= Density/ Molar mass

Pure solids and liquids have constant densities and molar masses at a constant temperature.

Because of this, they have constant molar concentrations, which means that they are not considered in the equilibrium constant for a heterogeneous reaction.

Learn more about the pure solids, liquids, and heterogenous mixture here,

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2 years ago

Consider the reaction: A (aq) ⇌ B (aq) at 253 K where the initial concentration of A = 1.00 M and the initial concentration of B

Ugo [173]

Answer:

The maximum amount of work that can be done by this system is -2.71 kJ/mol

Explanation:

Maximum amount of work denoted change in gibbs free energy $(\Delta G)$ during the reaction.

Equilibrium concentration of B = 0.357 M

So equilibrium concentration of A = (1-0.357) M = 0.643 M

So equilibrium constant at 253 K, $K_{eq}= \frac{[B]}{[A]}$

[A] and [B] represent equilibrium concentrations

$K_{eq}=\frac{0.357}{0.643}=0.555$

When concentration of A = 0.867 M then B = (1-0.867) M = 0.133 M

So reaction quotient at this situation, $Q=\frac{0.133}{0.867}=0.153$

We know, $\Delta G=RTln(\frac{Q}{K_{eq}})$

where R is gas constant and T is temperature in kelvin

Here R is 8.314 J/(mol.K), T is 253 K, Q is 0.153 and $K_{eq}$ is 0.555

So, $\Delta G=8.314\times 253\times ln(\frac{0.153}{0.555})mol/K$

= -2710 J/mol

= -2.71 kJ/mol

4 0

3 years ago

Can some answer this please?

marysya [2.9K]

Answer:

Explanation:

it's b because I just went over that frome my class and got it correct

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3 years ago

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Answer:

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