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3 years ago

Based on the activity series provided, which reactants will form products?

Chemistry

2 answers:

CaHeK987 [17]3 years ago

8 0

Idk man, but people are saying its Br2 + NaCl

hope it helped you

vovangra [49]3 years ago

7 0

Answer: CuI₂ + Br₂

Explanation:

1) The activity series F > Cl > Br > I means that F is the most active and I is the least active of those four elements (the halogens, group 17 in the periodic table).

The activity is a measure of how eager is an element to react compared to other elements in the series in a single replacement reaction.

2) Choice 1: CuI₂ + Br₂

Since the activity of Br is higher than that of I, Br will react with CuI₂, displacing I, which will be left alone, as per this chemical equation:

CuI₂ + Br₂ → CuBr₂ + I₂

Being I less active than Br, it cannot displace Br in CuBr₂.

3) Choice 2: Cl₂ + AlF₃

Being Cl less active than F, the former will not displace the latter, and the reaction will not proceed.

4) Choice 3: Br₂ + NaCl

Again, being Br less active than Cl, the former will not displace the latter, and the reaction will not proceed.

5) Choice 4: CuF₂ + I₂

Once more, being I less active than F, the former will not displace the latter, and the reaction will not proceed.

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4. Cuál es la aplicación de las reacciones de óxido-reducción en la vida diaria y en la industria?

12345 [234]

Answer:

Dentro de las aplicaciones de la óxido-reducción se pueden encontrar:

La obtención del aluminio a partir de la alúmina y la electrolisis.
La obtención de cloro, hidrógeno e hidróxido de sodio a partir del cloruro de sodio y la electrolisis.
La combustión interna de un motor a gasolina u otro combustible fósil.
Las termoeléctricas, las cuales para generar energía realizan combustión de carbón.
La galvanoplastia, donde para evitar la corrosión de un metal se recubre con otro metal más resistente, por ejemplo: el recubrimiento del acero con zinc.
La pilas o baterías de las cuales se obtiene energía química.

Explanation:

Como puedes ver en la respuesta, la óxido-reducción tiene diversas aplicaciones en la vida moderna, desde todos los tipos de combustión los cuales sirven para brindar energía o movilizarte, hasta todas las funciones que se le ha dado a la electrolisis y a la obtención de la energía por medios químicos, incluso se puede considerar una aplicación de la óxido-reducción la incorporación de antioxidantes en los alimentos, los cuales disminuyen la velocidad de descomposición de los mismos.

3 0

4 years ago

From the nature of the universe (that's if there is only one) to the purpose of dreams, there are lots of things we still don't

Flauer [41]

Answer:

Explanation: What is the universe made of?

Astronomers face an embarrassing conundrum: they don’t know what 95% of the universe is made of. Atoms, which form everything we see around us, only account for a measly 5%. Over the past 80 years it has become clear that the substantial remainder is comprised of two shadowy entities – dark matter and dark energy. The former, first discovered in 1933, acts as an invisible glue, binding galaxies and galaxy clusters together. Unveiled in 1998, the latter is pushing the universe’s expansion to ever greater speeds. Astronomers are closing in on the true identities of these unseen interlopers.

2 How did life begin?

Four billion years ago, something started stirring in the primordial soup. A few simple chemicals got together and made biology – the first molecules capable of replicating themselves appeared. We humans are linked by evolution to those early biological molecules. But how did the basic chemicals present on early Earth spontaneously arrange themselves into something resembling life? How did we get DNA? What did the first cells look like? More than half a century after the chemist Stanley Miller proposed his “primordial soup” theory, we still can’t agree about what happened. Some say life began in hot pools near volcanoes, others that it was kick-started by meteorites hitting the sea.

3 Are we alone in the universe?

science 3

Perhaps not. Astronomers have been scouring the universe for places where water worlds might have given rise to life, from Europa and Mars in our solar system to planets many light years away. Radio telescopes have been eavesdropping on the heavens and in 1977 a signal bearing the potential hallmarks of an alien message was heard. Astronomers are now able to scan the atmospheres of alien worlds for oxygen and water. The next few decades will be an exciting time to be an alien hunter with up to 60bn potentially habitable planets in our Milky Way alone.

6 0

3 years ago

A gas sample is held at constant pressure. The gas occupies 2.97 L of volume when the temperature is 21.6°C. Determine the tempe

STALIN [3.7K]

Answer:

339.2K

Explanation:

Using Charles law equation;

V1/T1 = V2/T2

Where;

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

V1 = 2.97 L

V2 = 3.42 L

T1 = 21.6°C = 21.6 + 273 = 294.6K

T2 = ?

Using V1/T1 = V2/T2

2.97/294.6 = 3.42/T2

Cross multiply

2.97 × T2 = 294.6 × 3.42

2.97T2 = 1007.532

T2 = 1007.532 ÷ 2.97

T2 = 339.236

The final temperature is 339.2K

5 0

3 years ago

if 28.5 g of calcium hydroxide is dissolved in enough water to make 185g of solution what is the percent by mass of calcium hydr

DENIUS [597]

The percent by mass of calcium hydroxide in the solution : 15.41%

<h3>Further explanation</h3>

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight/volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

Mass of solute (Ca(OH₂-Calcium hydroxide) : 28.5

Mass of solution = 185 g

$\tt \%mass=\dfrac{mass~solute}{mass~solution}\times 100\%\\\\\%mass=\dfrac{28.5~g}{185~g}\times 100\%\\\\\%mass=15.41\%$

6 0

3 years ago

During studies of the reaction below,

Leni [432]

Answer: The percent yield of the nitrogen gas is 11.53 %.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NO:

Given mass of NO = 11.5 g

Molar mass of NO = 30 g/mol

Putting values in equation 1, we get:

$\text{Moles of NO}=\frac{11.5g}{30g/mol}=0.383mol$

For $N_2O_4$ :

Given mass of $N_2O_4$ = 102.1 g

Molar mass of $N_2O_4$ = 92 g/mol

Putting values in equation 1, we get:

$\text{Moles of }N_2O_4=\frac{102.1g}{92g/mol}=1.11mol$

For the given chemical reactions:

$2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)$ ......(2)

$N_2H_4(l)+2N_2O_4(l)\rightarrow 6NO(g)+2H_2O(g)$ .......(3)

Calculating the experimental yield of nitrogen gas:

By Stoichiometry of the reaction 3:

6 moles of NO is produced from 2 moles of $N_2O_4$

So, 0.383 moles of NO will be produced from = $\frac{2}{6}\times 0.383=0.128mol$ of $N_2O_4$

By Stoichiometry of the reaction 2:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 0.128 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 0.128=0.384mol$ of nitrogen gas

Now, calculating the experimental yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 0.384 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$0.384mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.384mol\times 28g/mol)=10.75g$

Calculating the theoretical yield of nitrogen gas:

By Stoichiometry of the reaction 2:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 1.11 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 1.11=3.33mol$ of nitrogen gas

Now, calculating the theoretical yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 3.33 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$3.33mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(3.33mol\times 28g/mol)=93.24g$

To calculate the percentage yield of nitrogen gas, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of nitrogen gas = 10.75 g

Theoretical yield of nitrogen gas = 93.24 g

Putting values in above equation, we get:

$\%\text{ yield of nitrogen gas}=\frac{10.75g}{93.24g}\times 100\\\\\% \text{yield of nitrogen gas}=11.53\%$

Hence, the percent yield of the nitrogen gas is 11.53 %.

7 0

3 years ago