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3 years ago

Which of the following combinations led to a precipitation reaction?

Chemistry

1 answer:

Pani-rosa [81]3 years ago

3 0

(3) Silver Nitrate and Aluminium Sulfate

<u>Explanation:</u>

Silver nitrate and aluminium sulphate reacts to form Aluminium nitrate and silver sulfate.

The reaction is shown below:

6AgNO₃ (aq) + Al₂(SO₄)₃ (aq) → 2Al(NO₃)₃ (aq) + 3Ag₂SO₄ (S)

From the balanced equation we can see that Ag₂SO₄ or silver sulfate is the precipitate formed.

Aluminium sulfate and silver nitrate both are chemical compounds. There are many reactions that uses net ionic equation like acid-base neutralization reaction, redox reaction, double displacement reaction etc.

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kolbaska11 [484]

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2 years ago

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A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

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3 years ago

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frosja888 [35]

Property of matter, I think.

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Select the correct answer.

solong [7]

I think it’s A but I’m not sure, if it’s wrong I’m sorry

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