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Arada [10]
3 years ago
15

What identifies an ion

Chemistry
1 answer:
bezimeni [28]3 years ago
7 0

Answer:

Identifying whether or not an element is an ion is a very simple process. Identify the charge of the element. ... The number of electrons is equal to the atomic number minus the charge of the atom. Refer to an element with either a positive or negative charge as an ion.

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How many moles of bromine atoms are in 2.60×10^2 grams of bromine?
AlladinOne [14]
Your answer is 3.25 moles of Bromine

4 0
3 years ago
Chloroform, CHCl3, reacts with chlorine, Cl2, to form carbon tetrachloride, CCl4, and hydrogen chloride, HCl. In an experiment 2
erastovalidia [21]

Answer:

Chloroform= limiting reactant

0.209mol of CCl4 is formed

And 32.186g of CCl4 is formed

Explanation:

The equation of reaction

CHCl3 + Cl2= CCl4 + HCl

From the equation 1 mol of

CHCl3 reacts with 1mol Cl2 to yield 1mol of CCl4

From the question

25g of CHCl3 really with Cl2

Molar mass of CHCl3= 119.5

Molar mass of Cl2 = 71

Hence moles of CHCl3= 25/119.5 = 0.209mol

Moles of Cl2 = 25/71 = 0.352mol

Hence CHCl3 is the limiting reactant

Since 1 mole of CHCl3 gave 1mol of CCl4

It implies that 0.209moles of CHCl3 will also give 0.209mol of CCl4

Mass of CCl4 formed = moles× molar mass= 0.209×154= 32.186g

6 0
3 years ago
What is the element 3O2 name​
bija089 [108]

Answer:

Triplet oxygen

Explanation:

Based on my research it is called Triplet Oxygen, if this is wrong I'm sorry

5 0
3 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
2 years ago
What are prevailing winds?
Tems11 [23]

Answer:

D) winds that blow in the same direction at a consistent speed

Explanation:

i took the quiz got it right so i know the answer please trust me i know this is right i promise with all my heart

7 0
3 years ago
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