Aldehydes and Ketones. Aldehydes and ketones are classes of organic compounds that contain a carbonyl (C=O) group.
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Answer: The value of the equilibrium constant Kc for this reaction is 3.72
Explanation:
Equilibrium concentration of 
= 
Equilibrium concentration of 
= 
Equilibrium concentration of 
= 
Equilibrium concentration of 
= 
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For the given chemical reaction:
The expression for is written as:
![K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E3%5Ctimes%20%5BH_2O%5D%5E1%7D%7B%5BHNO_3%5D%5E2%5Ctimes%20%5BNO%5D%5E1%7D)
Thus the value of the equilibrium constant Kc for this reaction is 3.72
Answer:
63.25 grams of CO₂
Explanation:
To convert from liters to grams, we first need to convert from liters to moles. To do this, we divide the liters by 22.4, the amount of liters of a gas per mole.
32.2 / 22.4
= 1.4375 moles of CO₂
Now we want to convert from moles to grams. To do this, we multiply the moles by the molar mass of CO₂. The total molar mass can be found on the periodic table by adding up the molar mass of carbon (12) and two oxygen (32).
12 + 32 = 44
Now we want to multiply the moles by the molar mass.
1.4375 • 44
= 63.25 grams of CO₂
This is your answer.
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Answer:
the answer of this question is true
Answer:
340 grams Ca₃P₂ (2 sig. figs.)
Explanation:
3Ca + 2P => Ca₃P₂
5.6 mole + excess => ? grams
Convert the 'known' to a coefficient of 1 by dividing all coefficients by 3.
=> Ca + 2/3P => 1/3Ca₃P₂
From the above, 1 mole of Ca => 1/3 mole Ca₃P₂
∴ 5.6 mole Ca in an excess of P => 1/3(5.6 mole) Ca₃P₂
=> 1.8666 mol Ca₃P₂ (calculator answer) ≅ 1.9 mol Ca₃P₂
=> 1.9 mole x 182 g Ca₃P₂/mol Ca₃P₂ = 339.73333 grams Ca₃P₂
≅ 340 grams Ca₃P₂ (2 sig. figs.)
