What is the balanced form of the chemical equation shown below?

Chemistry

1 answer:

Svetradugi [14.3K]2 years ago

3 0

Answer:

Explanation:

Double Displacement reaction

Both sides are balanced with option D

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The diagrams show the arrangements of carbon atoms in diamond and in graphite. Compare a use of diamond with a use of graphite,

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Answer:

Explanation:

this is the diagram

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2 years ago

Saturated aqueous sodium chloride(d=1.2g/mL) is added to the following mixtures in order to dry the organic layer. Which layer i

natali 33 [55]

(a) at the bottom - high density organic compound dissolved in methylene chloride

(b) at the bottom - saturated aqueous sodium chloride

Explanation:

Sodium chloride is dissolved in water while the organic compounds are dissolved in methylene chloride. After mixing the two solutions two layers will form because water (polar molecule) will not mix with the methylene chloride (nonpolar molecule).

The layer with higher density will be at the bottom.

(a) saturated aqueous sodium chloride (d = 1.2 g/mL) - upper layer

high density organic compound dissolved in methylene chloride (d = 1.4 g/mL) - bottom layer

(b) saturated aqueous sodium chloride (d = 1.2 g/mL) - bottom layer

low density organic compound dissolved in methylene chloride (d = 1.1 g/mL) - upper layer

Learn more about:

liquids with different densities

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2 years ago

Se sabe que 10 g de calcio reaccionan con 4 g de oxígeno para obtener 14 g de óxido de calcio. Indica la cantidad de óxido de ca

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Answer:

Si se usan 50 gramos de calcio y óxigeno, se obtienen 70 gramos de óxido de calcio.

Explanation:

Hola,

En este caso, la reacción llevada a cabo es:

$2Ca+O_2\rightarrow 2CaO$

De este modo si asumimos el ejemplo dado, 50 gramos de calcio, cuya masa atómica es 40 g/mol y 50 g de oxígeno, cuya masa atómica como gas diatómico es 32 g/mol, antes de calcular los gramos de óxido de calcio producidos, debemos identificar el reactivo límite. Así, calculamos las moles de calcio disponibles en 50 g:

$mol_{Ca}^{disponible}=50gCa*\frac{1molCa}{40gCa} =1.25molCa$

Y también las moles de calcio consumidas por los 50 g de oxígeno, utilizando su relación molar 2:1:

$mol_{Ca}^{consumidas\ por\ O_2}=50gO_2*\frac{1molO_2}{32gO_2} *\frac{2molCa}{1molO_2} =3.125molCa$

Por lo tanto, hay menos calcio disponible que el que consume el oxígeno, por lo que el calcio esel reactivo límite. Ahora, con este, calculamos los gramos de óxido de calcio, cuya masa molar es 56 g/mol, que se producen:

$m_{CaO}=1.25molCa*\frac{2molCaO}{2molCa}* \frac{56gCaO}{1molCaO}\\ \\m_{CaO}=70gCaO$