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Lubov Fominskaja [6]
2 years ago
6

Cleaning PDB from the test tube after the experiment could be a long and arduous process. Describe the procedure that you will u

se to clean PDB (and the unknown solid) from your test tube after your experiments are complete.
Chemistry
1 answer:
fenix001 [56]2 years ago
6 0

The unknown solid is removed by rinsing with water while the PDB is removed by rinsing with alcohol.

<h3>What are solvents?</h3>

Solvents are substances which dissolve other substances to form solutions or mixtures.

Water is used as a solvent for inorganic substances while organic solvents such as alcohol or kerosene are used as solvents for organic substances such as PDB.

Alcohol is added to the test tube containing PDB, which then dissolves. The tube is rinsed with more alcohol until all the PDB dissolves.

Therefore, the unknown solid is removed by rinsing with water while the PDB is removed by rinsing with alcohol.

Learn more about solvents at: brainly.com/question/12665236

#SPJ2

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Write a balanced net ionic equation for each of the following aqueous metathesis reactions. Classify each reaction as a neutrali
melamori03 [73]

Answer:

See explanation

Explanation:

A. This is a neutralization reaction.

Molecular equation;

HBr(aq) + CsOH(aq) ---------> CsBr(aq) + H20(l)

Complete ionic equation;

H^+(aq) + Br^-(aq) + Cs^(aq) + OH^-(aq)   --------> Cs^+(aq) + Br^- + H20(l)

Net ionic equation;

H^+(aq) + OH^-(aq)   -------->  H20(l)

B. This is a gas forming reaction;

H2SO4(aq) + Na2CO3(aq) ------->Na2SO4(aq) + H2O(l) + CO2(g)

Complete ionic equation;

2H^+(aq) + SO4^-(aq) + 2Na^+(aq) + CO3^2-(aq)  ------->2Na^+(aq) + SO4^-(aq) + H2O(l) + CO2(g)

Net ionic equation;

2H^+(aq) + CO3^2-(aq)  -------> + H2O(l) + CO2(g)

C. This a precipitation reaction

Molecular equation;

CdCl2(aq) + Na2S(aq) ------->CdS(s) + 2NaCl(aq)

Complete ionic equation;

Cd^2+(aq) + 2Cl^-(aq) + 2Na^+(aq) + S^2-(aq) ---------> CdS(s) + 2Na^+(aq) +  2Cl^-(aq)

Net ionic equation;

Cd^2+(aq) +  S^2-(aq) ---------> CdS(s)

8 0
3 years ago
Which term describes a trace, print, or remain of an organism preserved over time in a rock?
kenny6666 [7]
B. Fossil

Why? Well skeleton seems like another answer BUT the definition of a fossil is an imprint of an organism on rock.

Example: A dinosaur presses it’s foot in dirt and it leaves a footprint or ‘fossil’ behind
5 0
3 years ago
The pH of a 0.25 M (aq) solution of hydrofluoric acid, HF, at 25C is 2.03. What is the value of Ka for HF? Pls show all mathemat
Snezhnost [94]

Answer:

Explanation:

b bhmj sjvkjwv

'

4 0
3 years ago
If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was
Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

n_{acid}=n_{base}=n_{salt}

Whereas the moles of the salt are computed as shown below:

n_{salt}=0.021L*0.68mol/L=0.01428mol

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

[salt]=0.01428mol/0.0276L=0.517M

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

C_6H_5NH_3^++H_2O\rightleftharpoons  C_6H_5NH_2+H_3O^+

Whose equilibrium expression is:

Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

2.326x10^{-5}=\frac{x^2}{0.517M}

Whereas x is:

x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

pH=-log(3.47x10^{-3})\\\\pH=2.46

Regards!

6 0
3 years ago
How many moles of a gas sample are in a 20.0 L container at 373 K and 203 kPa? The gas constant is 8.31 L−kPa/mol−K. A)0.33 mole
12345 [234]

Answer:

Option (C) 1.30 moles

Explanation:

The following data were obtained from the question:

Volume (V) = 20L

Temperature (T) = 373K

Pressure (P) = 203 kPa

Gas constant (R) = 8.31 L.kPa/mol.K.

Number of mole (n) =...?

The number of mole of the gas in the container can obtained by applying the ideal gas equation as illustrated below:

PV = nRT

Divide both side by RT

n = PV /RT

n = 203 x 20 / 8.31 x 373

n = 1.30 mole.

Therefore, 1.30 mole of the gas is present in the container.

4 0
3 years ago
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