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3 years ago

Hurrrry help me give me a experiment I can do at home that will be well conducted hellllp hurry answer

Chemistry

1 answer:

maksim [4K]3 years ago

6 0

Tornado in a bottle.
Rainbow in a glass.
Gooey slime.
Pasta rocket.
Homemade lava lamp.
Instant ice.
Ferromagnetic fluid.
Baking soda volcano.

You can search them up for more information all of them are pretty easy to do and if you do it right there is not much that can go wrong.

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A sample is decomposed in the laboratory and found that it contains 52.2% carbon (C), 13.0% hydrogen (H), and 34.8% oxygen (O).

Andrej [43]

The empirical formula would be C2H6O

7 0

4 years ago

Calculate the no. of each atoms presents in 10 grams of calcium carbonate.

raketka [301]

Molar Mass of Calcium carbonate:-

$\\ \sf\longmapsto CaCO_3$

$\\ \sf\longmapsto 40u+12u+3(16u)$

$\\ \sf\longmapsto 52u+48u$

$\\ \sf\longmapsto 100u$

$\\ \sf\longmapsto 100g/mol$

Given Mass=10g

$\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}$

$\\ \sf\longmapsto No\:of\;moles=\dfrac{10}{100}$

$\\ \sf\longmapsto No\:of\:moles=0.1mol$

Now

$\boxed{\sf No\:of\:molecules=No\;of\:moles\times Avagadro\: No}$

$\\ \sf\longmapsto 0.1\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 6.023\times 10^{22}molecules$

8 0

3 years ago

You need to make an aqueous solution of 0.182 M aluminum sulfate for an experiment in lab, using a 250 mL volumetric flask. How

Ksju [112]

Answer:

Hence, 15.99 g of solid Aluminum Sulfate should be added in 250 mL of Volumetric flask.

Explanation:

To make 0.187 M of Aluminum Sulfate solution in a 250 mL (0.250 L) Volumetric flask

The molar mass of Aluminum Sulfate = 342.15 g/mol

Using the molarity formula:-

Molarity = Number of moles/Volume of solution in a liter

Number of moles = Given weight/ molar mass

Molarity = (Given weight/ molar mass)/Volume of solution in liter

0.187 M = (Given weight/342.15 g/mol)/0.250 L

Given weight = 15.99 g

8 0

3 years ago

Gaseous methane (CH4) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H20). What

Mariana [72]

Answer:

Theoretical yield = 3.51 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $CH_4$ :-

Mass of $CH_4$ = 1.28 g

Molar mass of $CH_4$ = 16.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.28\ g}{16.04\ g/mol}$

$Moles_{CH_4}= 0.0798\ mol$

For $O_2$ :-

Mass of $O_2$ = 10.1 g

Molar mass of $O_2$ = 31.998 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{10.1\ g}{31.998\ g/mol}$

$Moles_{O_2}= 0.3156\ mol$

According to the given reaction:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

1 mole of methane gas reacts with 2 moles of oxygen gas

0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas

Moles of oxygen gas = 0.1596 moles

Available moles of oxygen gas = 0.3156 moles

<u>Limiting reagent is the one which is present in small amount. Thus, $CH_4$ is limiting reagent. </u>

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction produces 1 mole of carbon dioxide.

0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.

Mole of carbon dioxide = 0.0798 mole

Molar mass of carbon dioxide = 44.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0798\ moles= \frac{Mass}{44.01\ g/mol}$

Mass of $CO_2$ = 3.51 g

<u> Theoretical yield = 3.51 g</u>

3 0

3 years ago

A chemist dissolved crystals of an unknown substance into water at room temperature. He found that 33 g of the substance can be

Inga [223]

The answer is: It's solubility

8 0

4 years ago

Read 2 more answers