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Vesna [10]
3 years ago
10

Iron (Fe) is classified as a metal while Silicon (Si) is classified as a metalloid. Which physical property do these two element

s most likely have in common
Chemistry
1 answer:
sp2606 [1]3 years ago
4 0
I’ll try to get someone to help you I got you bro I’m gonna go find somebody got you
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What is a colligative property?
V125BC [204]
<span>Colligative properties are properties of solutions that depend on the number of molecules [or ions] in a given volume of solvent and not on the properties (e.g. size or mass) of the compound. Colligative properties include: lowering of vapor pressure; elevation of boiling point; depression of freezing point and osmotic pressure.</span>
6 0
3 years ago
What will lithium and bromine make when they bond
Rus_ich [418]

Answer:

they will form a compound

5 0
3 years ago
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Draw the Lewis structure for BrCl3. What are the approximate bond angles about the central atom?
Korvikt [17]

Answer:

Explanation:

BrCl₃ is an interhalogen compound with a hybridization of sp³d. The approximate bond angles can be predicted from the structure (attached below). Although, the lewis structure might be predicted to be trigonal bipyramidal from the structure, it is however a T-shaped geometry because of it's two lone pairs.

Also, from the structure attached, it can be predicted that the approximate bond angles about the central atom is 120° (360 ÷ 3) since each of the three chlorine atoms is equally spaced about the central atom.

7 0
2 years ago
Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s
NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

3 0
3 years ago
Solid aluminum metal and diatomic bromine liquid react spontaneously to form a solid product. give the balanced chemical equatio
kaheart [24]
Aluminum has a chemical formula of Al, while diatomic bromine has a chemical formula of Br₂. The balanced chemical reaction is shown below:

<em>2 Al (s) + 3 Br₂ (l) → 2 AlBr₃ (s)</em>

The solid product is called Dibromoaluminum. The stoichiometric coefficients are used to balance the reaction to obey the Law of Conservation of Mass.
3 0
3 years ago
Read 2 more answers
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