Answer:
A/1. 10.9 mol O2
Explanation:
583 g x 1 mol SO3 x 3 mol O2 /
80.057 g mol SO3 x 2 mol SO3
- You just need to find molar mass of SO3, which is 80.057 g.
- Everything else came from formula. Further explanation...
- Always start with what they give, such as 583 g. Then find 1 mol of what is being produced, in this it is SO3. We already found this because we did molar mass above. Next. find how many moles of what they want, which is O2. Look in equation and you can see 3 mol in from of O2. Next, do the same for SO3 and you can find 3 mol in front of that. Lastly, just do the math.
- If you need a further explanation or more help on any problems I would be happy to help, just let me know.
<span>They gain energy. In condensation, a gas becomes a liquid when this energy is removed (the cooling process).</span>
Answer: Superscript 32 subscript 15 upper P right arrow superscript 32 subscript 16 upper S plus superscript 0 subscript negative 1 e.
Explanation:
Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.
The general representation of beta minus emission is :
The representation of beta decay of phosphorous- 32 :
Superscript 32 subscript 15 upper P right arrow superscript 32 subscript 16 upper S plus superscript 0 subscript negative 1 e.
The percentage error that would be introduced in a 35.0 mL sample of liquid is 2.86%
<h3>Data obatined from the question</h3>
From the question given, the following data were obtained:
- Absolute error = 1 mL
- True measurement = 35 mL
- Perecentage error =?
<h3>How to determine the percentage error</h3>
The percentage error that would be introduced can be obtained as illustrated below:
Percentage error = (Absolute error / True measurement) × 100
Percentage error = (1 / 35) × 100
Percentage error = 2.86%
Thus, percentage error that would be introduced is 2.86%
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The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
The bond energies data is given as follows:
BE for C≡O = 1072 kJ/mol
BE for Cl-Cl = 242 kJ/mol
BE for C-Cl = 328 kJ/mol
BE for C=O = 766 kJ/mol
The enthalpy change for the reaction is given as :
ΔHr×n = ∑H reactant bond - ∑H product bond
ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )
ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )
ΔHr×n = 1314 - 1422
ΔHr×n = - 108 kJ
Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
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