Answer:
See below!
Explanation:
For the chemical formula, you need to have enough of each atom so that the charge is zero.
Aluminum has a +3 charge, and fluorine has a -1 charge. Since the charge has to be zero, you need three fluorines, giving you AlF₃.
Barium has a +2 charge, and oxygen has a -2 charge. Since the charges are equal in magnitude but opposite in sign, you only need one of each atom giving you BaO.
The name of the ionic compound will be the metal and then the nonmetal. When putting the nonmetal in, change the ending to "-ide". For example "chlorine" would be "chloride.
CaCl₂ ==> calcium chloride
Ga₂S₃ ==> gallium sulfide
K₃N ==> potassium nitride
AlF₃ ==> aluminum fluoride
BaO ==> barium oxide
Spectroscopy be used to distinguish between the following is the compound B has a peak at 3200 – 3500 cm⁻¹ in its IR spectrum.
<h3>What is spectroscopy?</h3>
Spectroscopy is the study of emission or absorption of light. It is used to study the structure of atoms and molecules.
The three types of spectroscopy are:
- atomic absorption spectroscopy (AAS)
- atomic emission spectroscopy (AES)
- atomic fluorescence spectroscopy (AFS)
Thus, the correct option is B, the compound B has a peak at 3200 – 3500 cm⁻¹ in its IR spectrum.
Learn more about spectroscopy
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Answer:
Solid Osmium transition metal reacts with Oxygen gas to produce solid Osmium tetroxide.
Os(s) + 2O₂(g) -> OsO₄(s)
Explanation:
Osmium tetroxide is another way of writing Osmium (VIII) oxide.
Leaving powdered osmium exposed to air in a room will slowly create osmium tetroxide at room temperature.
Similarly, osmium tetroxide vapor will readily be released from a liquid solution at room temperature.
Black hole, or a singularity contained within an event horizon through which no light can escape.
The molality of a solute is equal to the moles of solute per kg of solvent. We are given the mole fraction of I₂ in CH₂Cl₂ is <em>X</em> = 0.115. If we can an arbitrary sample of 1 mole of solution, we will have:
0.115 mol I₂
1 - 0.115 = 0.885 mol CH₂Cl₂
We need moles of solute, which we have, and must convert our moles of solvent to kg:
0.885 mol x 84.93 g/mol = 75.2 g CH₂Cl₂ x 1 kg/1000g = 0.0752 kg CH₂Cl₂
We can now calculate the molality:
m = 0.115 mol I₂/0.0752 kg CH₂Cl₂
m = 1.53 mol I₂/kg CH₂Cl₂
The molality of the iodine solution is 1.53.
