Answer:
The kind of ionic compound formed is MX2.
Explanation:
Element X electron configuration is represented as [core] ns2np5. The group in the periodic table this element belong to is group 7A. The element group is called the halogen family. Element X cannot be stated specifically, because the number is represented with n. Element X will behave as an anions when it react with a metal(cations). Element X has a charge of -1. The element X will gain electron when it bond with a metal. Element X is a non metal . Elements in this group are fluorine, chlorine, bromine, iodine , astatine, and tennessine . The element X have 7 valency electrons.
Element M electronic configuration is represented as [core]ns2. The group in the periodic table this element belong to is group 2A . The element group is called the alkaline earth metals family . Element M will behave as a cation when it bond with a non metal. Element M is a metal , therefore it will likely lose electron to form cations during bonding . The charge of element M is 2+. Element M is positively charged. Elements that belong to this group includes beryllium, magnesium, calcium, strontium, barium and radium. Element M has 2 valency electrons.
The reaction between this 2 ions will likely form an ionic compound . The element M is the cations while the element X is the anions. The element M will lose 2 electron while 2 atoms of element X will gain 2 electrons.Element M will lose 2 electron to attain a stable configuration while 2 atoms of element X will gain a single electron each to attain a stable configuration.
M²+ and F- . This will form MX2 when you cross multiply the charge. The kind of ionic compound formed is MX2.
Convert the child weight (37.3 pounds) to kilograms
37.3 lb x 0.453 kg /1lb = "A kg"
multiply the dose (9.00mg/kg) by the weight of the child to find how much you need to give him
A kg * 9.00 mg/1kg = "B mg"
calculate the mL of suspension dividing the "B mg" by the concentration of the suspension 60.0 mg/mL
B mg * 1mL/ 60.0 mg = C mL <span>oxcarbazepine</span>
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Titanium tribromide, titanium (III) bromide, or titanous bromide.
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Answer: The answer to the first one is the second option and the answer for the second one is the first option.
Explanation:
