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3 years ago

The____ cation will precipitate after the_ step. Then the _cation will precipitate after the____ step.

Chemistry

1 answer:

Novay_Z [31]3 years ago

8 0

Answer:

f. Sn^4+

c. second

e. Al^3+

d. third

Explanation:

This question comes from a quantitative analysis showing the flowchart of a common scheme for identifying cations.

Now, from the separation scheme, Let's assume that Sn⁴⁺ & Al³⁺ were given; Then, Yes, the separation will work.

However, there will be occurrence of precipitation after the 1st step1.

So, the Sn⁴⁺ cation will precipitate after the second step. Then the Al³⁺ cation will precipitate after the third step.

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Cryolite (Na3AlF6) is used in the production of aluminum from its ores. It is made by the reaction 6 NaOH 1 Al2O3 1 12 HF 8n 2 N

kap26 [50]

Answer: The mass of cryolite produced in the reaction is 1181.8 g

Explanation:

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

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

Given mass of aluminium oxide = 287 g

Molar mass of aluminium oxide = 102 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium oxide}=\frac{287g}{102g/mol}=2.814mol$

The given chemical reaction follows:

$6NaOH+Al_2O_3+12HF\rightarrow 2Na_3AlF_6+9H_2O$

By Stoichiometry of the reaction:

1 mole of aluminium oxide produces 2 moles of cryolite

So, 2.814 moles of aluminium oxide will produce = $\frac{2}{1}\times 2.814=5.628mol$ of cryolite

Now, calculating the mass of cryolite by using equation 1:

Moles of cryolite = 5.628 moles

Molar mass of cryolite = 210 g/mol

Putting values in equation 1, we get:

$5.628mol=\frac{\text{Mass of cryolite}}{210g/mol}\\\\\text{Mass of cryolite}=(5.628mol\times 210g/mol)=1181.8g$

Hence, the mass of cryolite produced in the reaction is 1181.8 g

5 0

3 years ago

Ethylene glycol (antifreeze) has a density of 1.11 g/cm3. What is the mass in grams of 385 mL of this liquid?

Makovka662 [10]

Cm^3 = mL

1.11 g/cm^3 = 1.11 g/mL

Density (g/mL) multiplied by volume (mL) will give us the mass (g)

1.11 g/mL * 385 mL = 427.35 g

And since we have 3 significant figures, we have 427. g.

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3 years ago

Calculate the following quantity: volume of 1.403 M copper(II) nitrate that must be diluted with water to prepare 575.2 mL of a

Juliette [100K]

Answer:

328 ml

Explanation:

We have given final volume =575.2 ml=0.575 L

Final concentration = 0.8012 M

We know that moles of copper(II) nitrate = final volume ×final concentration =0.8012×.0575=0.4606 moles

We have to find initial volume

So initial volume $=\frac{moles\ of \ copper(II)\ nitrate}{initial\ concentration}=\frac{0.4606}{1.403}=0.328L=328ml$

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4 years ago

What is one way to separate a mixture?

Dmitry_Shevchenko [17]

1. Substances in a mixture can be separated using different methods, for example distillation (separation of two liquids).
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3 years ago

What are some of the names given to the Moon by the Algonquin?

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4 years ago

Read 2 more answers

The________ cation will precipitate after the_________ step. Then the _________cation will precipitate after the________ step.

The____ cation will precipitate after the_ step. Then the _cation will precipitate after the____ step.