hydrobromic acid is added to tin metal express your answer as a balanced chemical equation. identify all of the phases in your answer. if no reaction occurs, simply write noreaction.

Answers

Answer 1

Tin metal is mixed with hydrobromic acid, creating a metal within acid situation. Therefore, in this instance, the metal is attempting to replace the hydrogen within it. (For more detail scroll down)

Is HBr a stronger acid or HCl?

The distance between the hydrogen and halogen atoms increases as the anion's size expands. It gets simpler to release H+ ions from hydracids as we move lower in the halogen group. Therefore, HI>HBr>HCl>HF is the sequence of acid strength.

Since tin is higher up, it will be able displace hydrogen from the acid to create the hydrogen gas. Tin can remove the hydrogen from the solution because it is more dense than hydrogen.

Sn(s) +2 HBr(aq) ----->SnBr2(aq) +H2(g)

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Related Questions

Beer's Law becomes nonlinear at high analyte concentrations
true or false

Answers

Answer:

True

Explanation:

Self-absorption occurs when the absorbing species in a sample interacts with other absorbing species in the same sample, leading to an overestimation of the absorbance and a deviation from the linear relationship described by Beer's Law. Therefore, at high analyte concentrations, the linear relationship breaks down and the law becomes nonlinear...

Which example is an biotic factor of an aquarium environment?

Answers

Answer: fish

Explanation: they are living organisms; they live in water

A biotic factor is a living organism that shapes its environment. In a freshwater ecosystem, examples might include aquatic plants, fish, amphibians, and algae.

Butylbenzene cannot be synthesized in good yield directly from benzene using a freidel-crafts alkylation. Propse an alternate synthesis of butylbenzene that does not use a freidel-crafts reaction.​

Answers

Provide an alternative method for producing that does not employ a Related trades alkylation and can be done in good yields by starting with benzene.

A Craft is what?

The trade of the carpenter. trades like sewing, carpentry, and ceramics; the craft of producing plays. Weaving, Quilting, Macramé, Lace-making, Knitting, and Tapestry Art.

A carpenter is what?

You would be involved with the planning, chopping, and construction of furniture as a carpenter utilising various types of wood. Working on a job site, at a client's house, or in a shop are all options. They build, set up, and fix wooden furniture, doors, and window frames, among other things.

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1.) How many grams of aluminum hydroxide can be produced from the
synthesis reaction of 24.3 g of aluminum oxide and 82.1 g of water?

Answers

Answer: 37.1 grams of aluminum hydroxide can be produced from the given amounts of aluminum oxide and water

Explanation:

The balanced chemical equation for the synthesis of aluminum hydroxide from aluminum oxide and water is:

Al2O3 + 3H2O -> 2Al(OH)3

From the equation, we can see that 1 mole of Al2O3 reacts with 3 moles of H2O to produce 2 moles of Al(OH)3.

We need to first calculate the number of moles of Al2O3 and H2O we have:

Number of moles of Al2O3 = mass / molar mass = 24.3 g / 101.96 g/mol = 0.238 moles

Number of moles of H2O = mass / molar mass = 82.1 g / 18.02 g/mol = 4.56 moles

From the balanced equation, we know that 1 mole of Al2O3 produces 2 moles of Al(OH)3. So, the number of moles of Al(OH)3 produced will be:

Number of moles of Al(OH)3 = (0.238 mol Al2O3) x (2 mol Al(OH)3 / 1 mol Al2O3) = 0.476 mol Al(OH)3

Finally, we can calculate the mass of Al(OH)3 produced:

Mass of Al(OH)3 = number of moles x molar mass = 0.476 mol x 78.0 g/mol = 37.1 g

Therefore, 37.1 grams of aluminum hydroxide can be produced from the given amounts of aluminum oxide and water.

2 Al + 6 HCl --> 2 AlCl3 + 3 H2

If 47.5 L of hydrogen gas is produced at STP, what mass in grams of Al was needed to react with excess HCl?

Answers

Answer:

To determine the mass of Al that was needed to produce 47.5 L of H2 gas at STP, we can use the stoichiometry of the balanced chemical equation to calculate the amount of Al required.

First, we need to convert the volume of H2 gas at STP to the corresponding amount in moles. At STP, 1 mole of gas occupies 22.4 L of volume. Therefore, 47.5 L of H2 gas is equal to 2.12 moles of H2 gas.

From the balanced chemical equation, we can see that 3 moles of H2 gas are produced for every 2 moles of Al consumed. Therefore, the amount of Al needed to produce 2.12 moles of H2 gas can be calculated as:

2.12 moles H2 x (2 moles Al / 3 moles H2) = 1.41 moles Al

Finally, we can use the molar mass of Al to convert moles of Al to its mass in grams. The molar mass of Al is 26.98 g/mol. Therefore, the mass of Al required can be calculated as:

1.41 moles Al x 26.98 g/mol = 38.1 g Al (rounded to one decimal place)

Therefore, approximately 38.1 grams of Al were needed to react with excess HCl to produce 47.5 L of H2 gas at STP.

Explanation:

In this balanced chemical equation, 2 moles of Al react with 6 moles of HCl to produce 2 moles of AlCl3 and 3 moles of H2. We are given the volume of H2 gas produced at STP and asked to calculate the mass of Al needed to react with excess HCl to produce this amount of H2 gas. To solve this problem, we used stoichiometry to relate the amount of H2 gas produced to the amount of Al required, and then used the molar mass of Al to convert from moles of Al to its mass in grams.

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