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pogonyaev
3 years ago
13

Al colocar unos trocitos de zinc en un tubo de ensayo, agregamos 2 mL de acido clorhídrico. El Zinc sustituye el hidrogeno del a

cido para formar una sal (ZnCl2) con el cloro Zn+2HCl-- ZnCl2 +H2 El nombre correcto de esta sal en la nomenclatura sistemática es
Chemistry
1 answer:
Rudik [331]3 years ago
8 0

Answer:

DICLORURO DE ZINC

Explanation:

La ecuación es:

Zn (s) + 2HCl (aq)  →  ZnCl₂ (aq) +H₂ (g)

El zinc reacciona en una reacción redox con el acido clorhídrico para formar una sal y liberar el hidrogeno gaseoso.

Se trata de una reacción redox porque el hidrógeno se reduce (disminuye el numero de oxidación, de -1 a 0, en su estado fundamental), mientras que el Zn se oxida (el numero de oxidación aumenta de 0 a +2).

Como la sal se forma con dos atomos de cloro, se denomina dicloruro por lo que el nombre correcto de la sal en nomenclatura sistemática es:

- DICLORURO DE ZINC

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A 151.5-g sample of a metal at 75.0°C is added to 151.5 g at 15.1°C. The temperature of the water rises to 18.7°C. Calculate the
Kryger [21]

Answer:

The specific heat capacity of the metal is 0.268 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 151.5 grams

The temperature of the metal = 75.0 °C

Temperature of water = 15.1 °C

The temperature of the water rises to 18.7°C.

The specific heat capacity of water is 4.18 J/°C*g

Step 2: Calculate the specific heat capacity of the metal

heat lost = heat gained

Q = m*c*ΔT

Qmetal = - Qwater

m(metal) * c(metal) * ΔT(metal) = m(water) * c(water) * ΔT(water)

⇒ mass of the metal = 151.5 grams

⇒ c(metal) = TO BE DETERMINED

⇒ΔT( metal) = T2 - T1 = 18.7 °C - 75.0 °C = -56.3 °C

⇒ mass of the water = 151.5 grams

⇒ c(water) = 4.184 J/g°C

⇒ ΔT(water) = 18.7° - 15.1 = 3.6 °C

151.5g * c(metal) * -56.3°C = 151.5g * 4.184 J/g°C * 3.6 °C

c(metal) = 0.268 J/g°C

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5 0
3 years ago
The observation that 20g of hydrogen gas always combines with 160g of oxygen gas to form 180g of water, even when there is more
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Answer:

The answer is most likely A. Definite proportions

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3 years ago
You wish to add 5 mg/l naocl as cl2 to a solution in a disinfection test, and you have a stock solution (household bleach) that
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Answer: -

0.1 ml of bleach should be added to each liter of test solution.

Explanation:-

Let the volume of bleach to be added is B ml.

Density of stock solution = 1.0 g/ml

Mass of stock solution = Volume of stock x density of stock

                                     = B ml x 1.0 g/ml

                                     = B g

Amount of NaOCl in this stock solution = 5% of B g

                                     = \frac{5}{100} x B g

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Now each test solution must be added 5 mg/l NaOCl.

Thus each liter of test solution must have 5 mg.

Thus 0.05 B g = 5 mg

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4 0
3 years ago
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4 years ago
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