Answer:
6.48 L
Explanation:
From the question,
Applying
PV/T = P'V'/T'......................... Equation 1
P = initial pressure of the helium balloon, V = Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.
make V' the subject of the equation
V' = PVT'/P'T......................... Equation 2
Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm
Substitute these values into equation 2
V' = (4.5×1×253)/(0.6×293)
V' = 1138.5/175.8
V' = 6.48 L
Answer:
See explanation
Explanation:
The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.
2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.
As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than 2-methylpropane.
Answer: 
is a reactant; it is present before the reaction occurs.
Explanation:
In a chemical reaction the chemical formulas written before the arrow are described as reactants as they react together to form products which are written after the arrow.
Thus and HCl are reactants here whereas 
, 
and 
are products.
You may find the Lewis dot structure of the CH₃-Br in the attached picture.
Explanation:
In the Lewis dot structure we represent the unpaired electrons using dots.
Bromide have one electron shared with one electron from the carbon to form a covalent bond, while the remaining electrons remains unpaired. We represent the six unpaired electrons of the bromide with dots.
Learn more about:
structure of organic compounds
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In Na2O, what is the oxidation state of oxygen? In Na2O oxidation state of Na is 1+
