All categories

3 years ago

1. What is the best method for removing water from a hydrated compound?

1 answer:

6 0

The traditional method is to heat the compound in an oven, on a hot plate or over a Bunsen burner. Waters of hydration are loosely bound in the compound and can be driven off at temperatures below the melting point of the compound. So the answer is heating.

You might be interested in

100 g of 20% salt solution is mixed with 200g of 10% salt solution. Find out the concentration of the resulting solution.

Hoochie [10]

Explanation:

The total mass is:

100 g + 200 g

= 300 g

The mass of salt is:

0.20 (100 g) + 0.10 (200 g)

= 20 g + 20 g

= 40 g

So the concentration is:

40 g / 300 g

≈ 13.33%

Round as needed.

7 0

3 years ago

If 13.2 kg of Al 2 O 3 ( s ) , 50.4 kg of NaOH ( l ) , and 50.4 kg of HF ( g ) react completely, how many kilograms of cryolite

Elodia [21]

Answer : The mass of cryolite produced will be, 54.38 kg

Solution : Given,

Mass of $Al_2O_3$ = 13.2 kg = 13200 g

Mass of $NaOH$ = 50.4 kg = 50400 g

Mass of $HF$ = 50.4 kg = 50400 g

Molar mass of $Al_2O_3$ = 101.9 g/mole

Molar mass of $NaOH$ = 40 g/mole

Molar mass of $HF$ = 20 g/mole

Molar mass of $Na_3AlF_6$ = 209.9 g/mole

First we have to calculate the moles of $Al_2O_3,NaOH$ and $HF$ .

$\text{ Moles of }Al_2O_3=\frac{\text{ Mass of }Al_2O_3}{\text{ Molar mass of }Al_2O_3}=\frac{13200g}{101.9g/mole}=129.54moles$

$\text{ Moles of }NaOH=\frac{\text{ Mass of }NaOH}{\text{ Molar mass of }NaOH}=\frac{50400g}{40g/mole}=1260moles$

$\text{ Moles of }HF=\frac{\text{ Mass of }HF}{\text{ Molar mass of }HF}=\frac{50400g}{20g/mole}=2520moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

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

From the balanced reaction we conclude that

The mole ratio of $Al_2O_3,NaOH$ and $HF$ is, 1 : 6 : 12

And, the ratio of given moles of $Al_2O_3,NaOH$ and $HF$ is, 129.54 : 1260 : 2520

From this we conclude that, $NaOH,HF$ is an excess reagent because the given moles are greater than the required moles and $Al_2O_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Na_3AlF_6$

From the reaction, we conclude that

As, 1 mole of $Al_2O_3$ react to give 2 mole of $Na_3AlF_6$

So, 129.54 moles of $Al_2O_3$ react to give $129.54\times 2=259.08$ moles of $Na_3AlF_6$

Now we have to calculate the mass of $Na_3AlF_6$

$\text{ Mass of }Na_3AlF_6=\text{ Moles of }Na_3AlF_6\times \text{ Molar mass of }Na_3AlF_6$

$\text{ Mass of }Na_3AlF_6=(259.08moles)\times (209.9g/mole)=54380.892g=54.38kg$

Thus, the mass of cryolite produced will be, 54.38 kg

5 0

3 years ago

Calculate the ph for each case in the titration of 50.0 ml of 0.150 m hclo(aq) with 0.150 m koh(aq). use the ionization constant

Lena [83]

<h3><u>Answer</u>;</h3>

pH =7.58

<h3><u>Explanation;</u></h3>

The KOH will react with the HClO to produce KClO. You will have a solution containing unreacted HClO and KOH. This is a buffer solution.

Equation:

HClO + KOH → KClO + H2O

HClO reacts with KOH in 1:1 molar ratio

Mol HClO in 50mL of 0.150M solution = 50/1000*0.150 = 0.0075 mol HClO

Mol KOH in 30mL of 0.150M solution = 30/1000*0.150 = 0.0045 mol KOH

These react to produce 0.0045 mol KClO and there is 0.0030 mol HClO unreacted

Volume of solution = 50mL + 30mL = 80mL = 0.080L

Molarity of HClO in solution = 0.0030/0.080 = 0.0375M

Molarity of KClO in solution = 0.0045/0.080 = 0.0562M

Using the Henderson-Hasselbalch equation we can calculate the pH;

pKa HClO = -log ( 4.0*10^-8) = 7.40

pH = pKa + log ([KClO]/[HClO])

pH = 7.40 + log( 0.0562/0.0375)

pH = 7.40 + log 1.50

pH = 7.40+ 0.18

pH = 7.58

7 0

4 years ago

Read 2 more answers

Match the following. 1. A mixture that does not have a uniform composition and the individual components remain distinct. polar

Ganezh [65]

Answer:

Explanation:

1. A mixture that does not have a uniform composition and the individual components remain distinct.

HETEROGENEOUS

An heterogeous mixture is a mixture with components in different phases.

2. A mixture that does have a uniform composition throughout and is always in the same state.

HOMOGENOUS

Homogenous mixtures have just one phase that is uniform all through.

3. A substance that will not dissolve in a solvent.

INSOLUBLE

When a solute cannot dissolve in a solvent to form a solution, we say it is insoluble.

4. A homogeneous mixture

SOLUTION

Solutions are made up of homogenous mixtures solute and solvent.

5. A molecule with no internal charge variation due to bonding.

NON-POLAR

Even distribution of charges especially between species whose electronegativity difference is 0 would lead to the formation of a non-polar compound. Here,

6. A molecule with an uneven distribution of charge due to unequal sharing of electrons during bonding

POLAR

Unequal sharing of electrons forms a polar compound. The more electronegative attracts the shared electron to itself and there is separation of charges. This leads to polarity.

7. A solution which has dissolved as much solute as it can at a particular temperature.

SATURATED

A saturated solution cannot dissolve more solute beacuse it contains enough solute as it can dissolve at a temperature.

8. A solution which is still able to dissolve solute.

UNSATURATED

An unsaturated solution is able to dissolve more solute.

8 0

3 years ago

Which symbol equation correctly describes the reaction:

larisa [96]

The answer is a I think

5 0

3 years ago

Read 2 more answers