If two gases react, compressing the gases generally increases the rate of reaction.
It states that a chemical reaction can occur the reacting particles must hit each other is called Collision theory.
A chemical reaction is a process in which the chemical bonds between atoms to break and reorganize, to form other new substances. In chemical reaction matter cannot be created or destroyed. It is called the law of conservation of mass.
Thus from the above conclusion we can say that If two gases react, compressing the gases generally increases the rate of reaction.
Learn more about the Collision Theory here: brainly.com/question/20628781
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What would happen to the entropy in the reaction 2O3(g) --> 3O2(g)?
It increases.
Answer:
Enthalpy and entropy. ... What would happen to the entropy in the reaction 2O3(g) 3O2(g)? It increases
Answer: The correct answer is Option b.
Explanation:
Water is the compound which is formed by the covalent bonding of hydrogen and oxygen atoms. The chemical formula for this compound is
Self-ionization of water is defined as the process in which 2 water molecules react together to produce hydroxide ion and hydronium ion.
The equation for the self ionization of water molecule follows:
By Stoichiometry of the reaction:
2 moles of water molecules react together to produce 1 mole of hydronium ion and 1 mole of hydroxide ion.
Hence, the correct answer is Option b.
The molecular formula of a compound with an empirical formula of CH₂O and a molar mass of 90 g/mol is C₃H₆O₃. This is found by dividing the compound's molar mass by the molar mass of the empirical formula, and using the ratio to multiply the subscripts in the empirical formula.
The molecular formula of a compound can be determined using the molar mass and the empirical formula given. The empirical formula for the compound is CH₂O, which has a molar mass of approximately 30 g/mol (12 g for C, 1 g for H, and 16 g for O). If the molar mass of the compound is 90 g/mol, we can find the ratio of the molar mass of the compound to the empirical formula by dividing 90 g/mol by 30 g/mol, which gives us 3.
This means the molecular formula of this compound is three times the empirical formula. So, to obtain the molecular formula, you multiply each subscript in the empirical formula by 3. If you do this for CH₂O, you get C₃H₆O₃ which is the molecular formula of the compound.
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Answer: Yes, lead ion will react with potassium iodide.
Explanation: When lead ion reacts with KI, a bright yellow colored precipitate is formed.
The reaction of these two follows:
Hence, ion reacts with KI to yield
This test is the confirmatory test for lead ion which is present in Group I.
is an endothermic process. A 6.1589-g sample of the solid is placed in an evacuated 4.000-L vessel at exactly 24°C. After equilibrium has been established, the total pressure inside is 0.709 atm. Some solid NH4HS remains in the vessel.
(a) What is the KP for the reaction?
(b) What percentage of the solid has decomposed?
(c) If the volume of the vessel were doubled at constant temperature, what would happen to the amount of solid in the vessel?
Answer:
(a) Kp = 0.126
(b) 48.1 %
(c) More of the solid will decompose, therefore it will decrease.
Explanation:
(a) We have here the equilibrium decomposition of NH₄HS, according to the equation:
NH₄HS (s) ⇄ NH₃ (g) + H₂S
with an equlibrium cosnstant, Kp, given by
Kp = pNH₃ x pH₂S
where pNH3 and pH₂S are the partial pressures of NH₃ and H₂S .
Since 1 mol NH₃ is produced for every 1 mol H₂S, it follows that the partial pressure of NH₃ is equal to the partial pressure of H₂S ( from ideal gas law the pressure is proportional to number of mol):
pNH₃ = pH₂S = 1/2 (0.709) atm = 0.355 atm
and
Kp = 0.355 x 0.355 = (0.3545) ² = 0.126
(b) To solve this part we need to do a calculation based on the stoichiometry of the reaction by calculating the number of moles, n, a partial pressure of 0.355 atm represent.
From the ideal gas law we can calculate it:
PV = nRT ⇒ n = PV/RT
where P = 0.355 atm, V = 4.000 L, T = ( 24 + 273 ) K, and R is the gas constant 0.08205 Latm/Kmol.
n = 0.355 atm x 4.000 L / 0.08205 Latm/Kmol x 297
n = 0.058 mol
Now we can relate this 0.058 mol of NH₃ ( or H₂S ) to the number of moles of moles of NH₄HS that must have decompesed. Since it is a 1: 1 ratio:
(1 mol NH₄HS / 1 mol NH₃ ) x 0.058 mol NH₃ = 0.058 mol NH₄HS
Having the molar mass of NH₄HS, 51.11 g/mol we can calculate its mass:
0.058 mol NH₄HS x 51.11 g/mol = 2.9644 g
Percentage decomposition is then equal to:
2.9644 / 6.1589 g x 100 g = 48.1 %
(c) If the volume of the vessel, effectively we are reducing the pressure by a half , and the system will react according to LeChatelier's principle by producing more gaseous products to reach equilibrium again. Therefore, more of the solid NH₄HS will decompose, and the amount of it will we reduced compared to the previous part.