A possible cause of a large percentage of error in an
experiment where MgO is produced from the combustion of magnesium would be not all of the Mg has
completely reacted. <span>
I hope this helps and if you have any further questions, please don’t hesitate
to ask again. </span>
Answer:
Part 1: - 1.091 x 10⁴ J/mol.
Part 2: - 1.137 x 10⁴ J/mol.
Explanation:
Part 1: At standard conditions:
At standard conditions Kp= 81.9.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(81.9)) = - 1.091 x 10⁴ J/mol.
Part 2: PICl = 2.63 atm; PI₂ = 0.324 atm; PCl₂ = 0.217 atm.
For the reaction:
I₂(g) + Cl₂(g) ⇌ 2ICl(g).
Kp = (PICl)²/(PI₂)(PCl₂) = (2.63 atm)²/(0.324 atm)(0.217 atm) = 98.38.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(98.38)) = - 1.137 x 10⁴ J/mol.
Answer:
The atomic number is the number of protons in the nucleus
Their lungs would try to expand to about 4 timed the normal volume which would force air into the various body tissues. this can cause a lung expansion injury and it could case air embolism. Air embolism is when air bubbles get trapped in blood vessels. This can lead to a blockage which will could be fatal.
To start, 1 cubic centimeter = 1 milliliter, so now you have 1.11g/mL.
Now multiply 1.11 by 387 to get the mass of antifreeze in grams, since the mL is canceled out.
387 mL x 1.11g/mL = 429.57 g
