If you would draw the Lewis structures of these atoms, you would see that A has 2 electron pairs and 2 lone electrons (that can bond). For B you’d see that you only have 1 electron that can form a bond. This means that 1 atom of A (2 lone electrons) can bond with 2 atoms of B. To know the kind of bond you have to know wether or not there will be a ‘donation’ of an electron from one atom to another. This happens when the number of electrons on one atoms is equal to the number of electrons another atom needs to reach the noble gas structure. As you can see, this is not the case here. This means that you get an AB2 structure with covalent character.
Answer:
7.50×10¯² lbs = 7.50×10¯² lbs × 453.952 g / 1 lbs
7.50×10¯² lbs = 34.0464 g.
Explanation:
From the question given, the following data were obtained:
1 lbs = 453.952 g
7.50×10¯² lbs =.?
Thus, we can obtain convert 7.50×10¯² lbs to grams as follow:
1 lbs = 453.952 g
Therefore,
7.50×10¯² lbs = 7.50×10¯² lbs × 453.952 g / 1 lbs
7.50×10¯² lbs = 34.0464 g
Therefore, 7.50×10¯² lbs is equivalent to 34.0464 g.
Answer:
if ice is added more salt can be dissolved ...........
The answer will be True to this question
Answer:
-4.7°C
Explanation:
The freezing point depression is a colligative property that can be calculated using the following expression.
ΔTf = Kf . b
where
ΔTf: depression in the freezing point
Kf: molal freezing point depression constant
b: molality
The moles of urea (solute) are:
The mass of solvent (X) is 500 g (0.500 kg).
The molality of the urea is:
ΔTf = Kf . b = (6.19 °C.kg/mol) × 0.34 mol/kg = 2.1°C
The freezing point of pure X is:
-6.8°C + 2.1°C = -4.7°C
