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3 years ago

How carbon dioxide prepared in lab?

Chemistry

1 answer:

Nana76 [90]3 years ago

7 0

Explanation:

In laboratory, carbon dioxide is not obtained from the air, carbon dioxide is prepared from carbonates by adding acids to the carbonates. In this experiment carbon dioxide is prepared by the reaction of marble with hydrochloric acid.

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Poor precision in scientific measurement may arise from

Arturiano [62]

Based on the experience of the responder, to correctly calculate measurements in real-world. Firstly is to avoid errors as much as possible. Errors are what makes your measurement invalid and unreliable. There are two types of error which is called the systematic error and the random error. Each error has different sources. Words that were mentioned –invalid and unreliable are very important key aspects to determine that your measure is truly accurate and consistent. Some would recommend using the mean method, doing three trials in measuring and getting their mean, in response to this problem.

7 0

4 years ago

45) George is making spaghetti for dinner. He places 4.01 kg of water in a pan and brings it to a boil.

lesya692 [45]

On adding salt.....The boiling temperature increases.....

So ∆t= KB * molality
=O.52*(58/58)/4
= O.52*1/4
= 0.13
So increase is 100+.13=100.13°c

5 0

3 years ago

If you have a solution that contains 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH, what is the mole fraction of

irakobra [83]

Answer:

0.22

Explanation:

Given, Mass of $C_{18}H_{21}NO_3$ = 46.85 g

Molar mass of $C_{18}H_{21}NO_3$ = 299.4 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$Moles= \frac{46.85\ g}{299.4\ g/mol}$

$Moles\ of\ C_{18}H_{21}NO_3= 0.1565\ mol$

Given, Mass of $C_{2}H_{5}OH$ = 125.5 g

Molar mass of $C_{2}H_{5}OH$ = 46.07 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$Moles= \frac{125.5\ g}{46.07\ g/mol}$

$Moles\ of\ C_{2}H_{5}OH= 0.5535\ mol$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ codeine=\frac {n_{codeine}}{n_{codeine}+n_{ethanol}}$

$Mole\ fraction\ of\ codeine=\frac{0.1565}{0.1565+0.5535}=0.22$

4 0

3 years ago

Which of the following statements best explains the observation that hydrogen fluoride has the highest boiling point of all the

olasank [31]

Answer:

c. HF can participate in hydrogen bonding.

Explanation:

The boiling points of substances often reflect the strength of the intermolecular forces operating among the molecules.

If it takes more energy to separate molecules of HF than of the rest of the hydrogen halides because HF molecules are held together by stronger intermolecular forces, then the boiling point of HF will be higher than that of all the hydrogen halides.

A particularly strong type of intermolecular attraction is called the hydrogen bond, which is a special type of dipole-dipole interaction between the hydrogen atom in a polar bond, such as N-H, O-H, or F-H, and an electronegative O, N, or F atom.

7 0

3 years ago

Hydrofluoric acid and Water react to form fluoride anion and hydronium cation, like this HF(aq) + H_2O(l) rightarrow F(aq) + H_3

maksim [4K]

Answer:

Kc = 1.09x10⁻⁴

Explanation:

HF = 1.62g

H₂O = 516g

F⁻ = 0.163g

H₃O⁺ = 0.110g

To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:

Kc = [H₃O⁺] [F⁻] / [HF]

Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:

[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M

[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M

[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M

Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]

7 0

3 years ago