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aleksley [76]
3 years ago
10

Can someone help me please? I’d appreciate it! Also please do not send me a link or delete your answer! Thanks

Chemistry
1 answer:
chubhunter [2.5K]3 years ago
5 0

Answer:

14.53ML

Explanation:

V1=218

V2=?

P2=15p1

USING BOYLE'S LAW

P1V1=P2V2

V2=P1V1/P2=P1(218ML)/15P1

=14.53ML

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Nitrogen monoxide and water react to form ammonia and oxygen, like this: (g)(g)(g)(g) Write the pressure equilibrium constant ex
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Answer:

Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}

Explanation:

First, we have to write the balanced chemical equation for the reaction. Nitrogen monoxide (NO) reacts with water (H₂O) to give ammonia (NH₃) and oxygen (O₂), according to the following:

NO(g) + H₂O(g) → NH₃(g) + O₂(g)

To balance the equation, we add the stoichiometric coefficients (4 for NH₃ and NO to balance N atoms, then 6 for H₂O to balance H atoms and then 5 for O₂ to balance O atoms):

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All reactants and products are in the gaseous phase, so the equilibrium constant is expressed in terms of partial pressures (P) and is denoted as Kp. The Kp is expressed as the product of the reaction products (NH₃ and O₃) raised by their stoichiometric coefficients (4 and 5, respectively) divided into the product of the reaction reagents (NO and H₂O) raised by their stoichiometric coefficients (4 and 6, respectively). So, the pressure equilibrium constant expression is written as follows:

Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}

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3 years ago
You have an aqueous solution of chromium(III) nitrate that you titrate with an aqueous solution of sodium hydroxide. After a cer
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The precipitate that is formed is chromium hydroxide, Cr(OH)_{3}

When more NaOH is added, the precipitate reacts with it which then results in the formation of a soluble complex ion:

Cr(OH)_{3}(s) + OH^{-}(aq) → Cr(OH)_{4} ^{-}(aq)

Cr(OH)_{4} ^{-} is soluble complex ion

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