State the coefficient required to correctly balance the following chemical equation:

Chemistry

1 answer:

frez [133]3 years ago

7 0

Answer:

2kcl+ fe= Fecl2+2K

so coefficient required be 2 for kcl and 2 forK

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Round off each number to the indicated number significant figures (sf):

qaws [65]

Answer :

(a) The given number in 2 significant figures will be, $3.6\times 10^3$

(b) The given number in 4 significant figures will be, $3.583\times 10^1$

(c) The given number in 3 significant figures will be, $2.24\times 10^1$

Explanation :

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

(a) The given number 00003554 converted into 2 significant figures.

In 00003554, there are 4 significant figures. Now we have to convert it into 2 significant figures.

The given number in 2 significant figures will be, $3.6\times 10^3$

(b) The given number 35.8348 converted into 4 significant figures.

In 35.8348, there are 6 significant figures. Now we have to convert it into 4 significant figures.

The given number in 4 significant figures will be, $3.583\times 10^1$

(c) The given number 22.4555 converted into 3 significant figures.

In 22.4555, there are 6 significant figures. Now we have to convert it into 3 significant figures.

The given number in 3 significant figures will be, $2.24\times 10^1$

7 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

3 years ago

Water is composed of one atom of oxygen and two atoms of

egoroff_w [7]

Answer:

<em>Hydrogen.</em>

Explanation:

You've probably seen " $H_{2}0$ " which is the formula for water. It means that there's 2 hydrogen atoms, and one oxygen atom, in one molecule of water.