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Luden [163]
3 years ago
8

A chemical that can superheat or give off poisonous vapors when it comes in contact with air or water is called

Chemistry
1 answer:
blagie [28]3 years ago
6 0
Radioactive ;-):-)

is that answer
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How many grams are in 2.5 moles of N2? <br><br> (2 decimal places)
Leokris [45]

Answer: How many grams are in 2.5 moles of N2?

Explanation: 1 mole is equal to 1 moles N2, or 28.0134 grams.

 One mole of N2 molecules would have a mass of 2 X 14.01 g = 28.02 g.

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3 years ago
12. Organelles allow eukaryotic cells to carry out more functions than prokaryotic cells.
deff fn [24]

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True po

Explaination :

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4 0
3 years ago
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Consider the following statements.
Vsevolod [243]

Answer: B and C only

Explanation:

3 0
3 years ago
I NEED HELP ASAP PLEASE
AnnyKZ [126]

One molecule of ammonia is composed of two atoms of nitrogen and three atoms of hydrogen. Option B.

<h3>What is an equation?</h3>

The term chemical equation has to do with the presentation of a chemical reaction on paper in a way that it can be easily understood. It is easy to write an equation to show what is going on in a reaction system.

Now we have the reactions as shown in the question. In this reaction which is the synthesis of ammonia and occurs industrially in the Haber process. The statement that is not true is that; one molecule of ammonia is composed of two atoms of nitrogen and three atoms of hydrogen. Option B.

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4 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
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