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Vera_Pavlovna [14]
3 years ago
10

What is this question in with the question tell me a funny joke cuz I'm having a bad day ​

Chemistry
2 answers:
NikAS [45]3 years ago
8 0

Im sorry, but I can not see the attachment... But I can try to help you cheer up! You shouldnt let anyone put you down, and certainly dont let life knock you down. Your strong, brave, and deserve happiness. Stay strong, stay brave, stay unique, stay you.

Leno4ka [110]3 years ago
6 0
<h2><u><em>Answer:</em></u></h2><h2><u><em>the one below is basically noble gas</em></u></h2><h2><u><em>For example, helium, neon, and argon are some of the individual atoms which are also known as noble gases and have very low chemical reactivity. ... Liquids are made of vibrating particles of matter that are held together by molecular bonds (e.g. atoms). The most common liquid you'll find is water</em></u></h2><h2><u><em>An element is a pure substance as well, because if we fill up a balloon with just helium gas, it will only contain helium atoms. On the periodic table of elements, helium is element number 2 , and so that should tell you that it's an element of course...</em></u></h2><h2><u><em>why didn't the skeleton cross the street?</em></u></h2><h2><em><u>"I dunno why"</u></em></h2><h2><em><u>because he didn't have the guts to do it</u></em></h2><h2><em><u>hahahah</u></em></h2><h2><em><u>mark brainliest plzs</u></em></h2>
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Consider the equilibrium reaction and its equilibrium constant expression. Br 2 ( g ) + 2 NO ( g ) − ⇀ ↽ − 2 NOBr ( g ) K = [ NO
zavuch27 [327]

Answer:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Explanation:

Hello,

In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

2 Br_2 ( g ) + 4 NO ( g ) \rightleftharpoons  4NOBr ( g )

The suitable equilibrium constant turns out:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Or in terms of the initial equilibrium constant:

K_2=K_1^2

Since the second reaction is a doubled version of the first one.

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