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2 years ago

Why is it important to use chemicals of high quality (purity) and exact(correct) concentrations in experiments?

Chemistry

2 answers:

marishachu [46]2 years ago

8 0

Answer:

To produce good or targeted result

Darya [45]2 years ago

6 0

It is important to use chemicals of high quality and correct concentrations in experiment so that we can obtain the predictable result in an experiment

Purity of substances/chemicals is very important when performing experiments in the laboratories, because it plays an important role in the determination of the chemical properties of substances which will have a direct effect on the result obtained from the experiment. likewise using the correct concentration of reagents

During experiments to obtain the best/predictable result in an experiment it is of best practice to make use of chemicals of high quality and also at the exact(needed) concentrations, that way the predictable result of the experiment will be obtained.

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Find the pH of the equivalence point(s) and the volume (mL) of 0.0372 M NaOH needed to reach the point(s) in titrations of(a) 42

Neko [114]

The pH of the equivalence points is 8.54

Concept of pH

A solution's acidity or alkalinity can be determined based on the concentration of hydrogen ions in the solution, or pH. Acidic aqueous solutions at 25 °C have a pH under 7, while basic or alkaline aqueous solutions have a pH above 7. Since the concentration of H3O+ is equal to the concentration of OH in pure water, a pH level of 7.0 at 25°C is referred to as "neutral". Strong bases may have a pH above 14, while very strong acids may have a pH below 14.

0.0520 M CH3COOH in 42.2 mL of moles is as follows:

2.194x103 mol CH3COOH = 0.0422L (0.0520mol / L)

that react with NaOH, resulting in:

NaOH + CH3COOH = CH3COO + Na+ + H2O

Thus, 1 mole of acetic acid and 1 mole of NaOH react.

As a result, 2.194x103 mol of NaOH are required to reach the equivalence point in volume:

To attain the equivalency point, 2.194x103 mol (1L / 0.0372mol) = 0.05899L 58.99mL of 0.0372 M NaOH

You will only have CH3COO at the equivalency point because it is in equilibrium with water, so:

H2O(l) + CH3COO(aq) CH3COOH(aq) + OH (aq)

A definition of equilibrium is:

Kb = 5.6x1010 = [OH] / [CH3COO] / [CH3COOH]

2.194x103mol of CH3COO has a molarity of (0.05899L + 0.0422L) = 0.02168M.

Therefore, equilibrium concentrations are:

[CH3COO]=0.02168M-X [CH3COOH]=X [OH]=X

5.6x1010 = [X] [X] / [0.02168M - X] converts to Kb.

1.214x1011 - 5.6x1010X = X2 X2 + 5.6x1010X - 1.214x1011 = 0 Finding the value of X:

False response; there are no negative concentrations. X: -3.48x106

As [OH] = X, [OH] = 3.484x106M, X is 3.484x106.

As 14 = pOH + pH pH = 8.54, pOH = -log [OH], or 5.46.

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2 years ago

Consider the following voltaic cell:(b) In which half-cell does oxidation occur?

damaskus [11]

At the anode, half-cell oxidation occurs in a voltaic cell.

<h3>Voltaic Cell Principle</h3>

A voltaic cell generates electricity due to the Gibbs free energy of spontaneous redox processes occurring inside the cell, which is the basis for the voltaic cell's operating principle.

Two half-cells plus a salt bridge make up the voltaic cell. An electrolyte-immersed metallic electrode is present on each side of the cell. These two half-cells are wired together to form a connection to a voltmeter.

<h3>Voltaic Cell Parts</h3>

Copper makes comprises the cathode of a photovoltaic cell. This electrode serves as the cell's positive terminal, where reduction takes place.
Anode: Zink metal makes up this electrode. It creates the cell's negative electrode, where oxidation takes place.
Oxidation and reduction are divided into two discrete parts in two half-cells.
Salt Bridge: It contains the electrolytes needed to finish the circuit in the voltaic cell.
The flow of electrons between the electrodes occurs via the external circuit.

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6 0

2 years ago

Electron configuration of ag

stepladder [879]

Silver/Electron configuration

[Kr] 4d¹⁰ 5s¹

Atomic number: 47

Symbol: Ag

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3 years ago

(will give brainliest) show your work. How many grams of Copper(I) nitrate, CuNO3 are required to produce 88.0 grams of aluminum

ValentinkaMS [17]

Based on the stoichiometry of the reaction, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

<h3>What is stoichiometry of a reaction?</h3>

The stoichiometry of a reaction is the molar ratio in which reactants combine to form products.

The stoichiometry of the reaction shows that 6 moles of copper (i) nitrate produces 2 moles of aluminium nitrate.

molar mass of Copper(I) nitrate, CuNO3 = 126 g

molar mass of aluminum nitrate, Al(NO3)3 = 213 g

88.0 g of aluminum nitrate, Al(NO3)3 = 88.0/213 moles = 0.413 moles

0.413 moles of Al(NO3)3 will be produced by 0.413 ×6/3 = 1.239 moles of CuNO3

Mass of 1.239 moles of CuNO3 = 1.239 × 126 = 156.114 g of CuNO3

Therefore, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

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Therefore, 156.114 g of CuNO3

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1 year ago

the purpose of the ____ system prevent foreign substances from entering the body and to attack foreign substances that are in th

Ratling [72]

Immune System is the answer.

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3 years ago

Why is it important to use chemicals of high quality (purity) and exact(correct) concentrations in experiments?​

Why is it important to use chemicals of high quality (purity) and exact(correct) concentrations in experiments?