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4 years ago

Which statements are true of precipitation reactions?

Chemistry

2 answers:

IgorC [24]4 years ago

8 0

The answer to this question would be A.

Leokris [45]4 years ago

6 0

Answer:

A C & D

Explanation:

its all that apply

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What cell part enables plant cell to carry out photossynthesis?

Luba_88 [7]

Chloroplasts aid in carrying out the process of photosynthesis to produce food for the plants.

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3 years ago

Read 2 more answers

A well-tested explanation for a set of observations or experimental results is a A. law. B. rule. C. theory. D. hypothesis.

faust18 [17]

c. a theory................

3 0

3 years ago

As with any combustion reaction, the products of combusting a hydrocarbon fuel (CxHy) with oxygen (O2) are carbon dioxide (CO2)

melisa1 [442]

Answer:

5 molecules of oxygen will be present 

Explanation:

In this reaction combustion is taking place hence there must be presence of oxygen in it.

We don't know the initial amount of carbon molecules in hydrogen molecules present hence the equation goes:-

$CxHy + O_2 => CO_2+ H_2O$

Mass of fuel 16.74 grams, Mass of carbon dioxide is 1198 gram and water is 0.0654 remaining mass of fuel is 16.70 g. Hence amount of fuel used up is 0.04 gram

 $0.04 + O => 0.1198 + 0.0654$ 

Weight of products is 0.1852 and weight of oxygen is 0.1452 

No of molecules will be $\frac {(0.1452 \times 6.023 \times 10^-^2^3)}{ (1.661 \times 10^-^2^4)} = 5$ molecules 

3 0

4 years ago

Number of grams of hydrogen than can be prepared from 6.80g of aluminum

Anastaziya [24]

Answer:

0.7561 g.

Explanation:

The hydrogen than can be prepared from Al according to the balanced equation:

2Al + 6HCl → 2AlCl₃ + 3H₂,

It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.

Firstly, we need to calculate the no. of moles of (6.8 g) of Al:

no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.

Using cross multiplication:

2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.

0.252 mol of Al need to react → ??? mol of H₂.

∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.

Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:

mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.

5 0

4 years ago

A sample of oxygen gas was collected via water displacement. since the oxygen was collected via water displacement, the sample i

mylen [45]

The sample of oxygen gas was collected through water displacement. So, the gas collected will be a mixture of oxygen and water vapor.

Given that the total pressure of the mixture of gases containing oxygen and water vapor = 749 Torr

Vapor pressure of pure water at $26.4^{0}C$ =25.81mmHg

= $25.81 mmHg*\frac{1 Torr}{1 mmHg} =25.81 Torr$

According to Dalton's law of partial pressures,

Total pressure = Partial pressure of Oxygen gas + Partial pressure of water

749 Torr = Partial pressure of Oxygen gas + 25.81 Torr

Partial pressure of Oxygen gas = 749 Torr - 25.81 Torr = 723.19 Torr

Therefore the partial pressure of Oxygen gas in the mixture collected will be 723.19 Torr

6 0

4 years ago