To reduce a haloalkane, it has to be eliminated first. This is because 2 bromopropanes are a saturated compound that can not be reduced by a saturated compound. Dehydrohalogenation of the haloalkanes into the form of propene and hydrogen bromide can lead to the elimination. The propene is then reduced to propane.
Answer:
–0.13 Pa.m²
Explanation:
From the question given above, the following data were obtained:
Measurement (Pa.mm²) = –1.3×10⁵ Pa.mm²
Measurement (Pa.m²) =?
We can convert from Pa.mm² to Pa.m² by doing the following:
1 Pa.mm² = 1×10¯⁶ Pa.m²
Therefore,
–1.3×10⁵ Pa.mm² = –1.3×10⁵ Pa.mm² × 1×10¯⁶ Pa.m² / 1 Pa.mm²
–1.3×10⁵ Pa.mm² = –0.13 Pa.m²
Thus, –1.3×10⁵ Pa.mm² is equivalent to –0.13 Pa.m².
Answer:
Constant pressure
Explanation:
At constant pressure,

At constant temperature,

1 mol of an ideal gas at STP has a volume of 22.71 L.
Let's compare the work done as it expands under each condition from an initial volume of 22.71 L.
Isobaric expansion

A plot of -w vs V₂ gives a straight line (red) with a constant slope of 100 J/L as in the diagram below (Note that w is work done on the system, so -w is the work done by the system).
\
Isothermal expansion

A plot of -w vs V₂ is a logarithmic curve. Its slope starts at 100 J/mol but decreases as the volume increases (the blue curve below).
Thus, the work done during an expansion at constant pressure is greater than if the system is at constant temperature.
B That the answer (hope this helps)
Answer:
See Below.
Explanation:
Recall that molarity is defined by moles of solute over liters of solution (mol/L).
Therefore, to make 90.0 mL of 2.0 M MgSO₄, we will need:

Convert this amount to grams. The molecular weight of MgSO₄ is 120.38 g/mol:

Therefore, to make the solution, we can add 22 grams of MgSO₄ into a graduated cylinder, then mix and dilute the solution with distilled water until we reach 90.0 mL.