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2 years ago

Why do we need to calibrate equipment

Chemistry

1 answer:

Vlad1618 [11]2 years ago

3 0

Answer:

to be precise in the calculations

Explanation: for example your cell phone has a gps inside it, the gps is responsible for providing direction data like a compass, if the equipment is not calibrated, the gps will not find the correct direction.

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A solution is made by dissolving 4.0 moles of sodium chloride (NaCl) in 2.05 kilograms of water. If the molal boiling point cons

kozerog [31]

Properties of a solution that depend only on the ratio of the number of particles of solute and solvent in the solution are known as colligative properties. For this problem, we use boiling point elevation concept.

ΔT(boiling point) = (Kb)mi
ΔT(boiling point) = (0.51 C-kg / mol )(4.0 mol / 2.05 kg ) (2)
ΔT(boiling point) = 1.99 C

T(boiling point) = 101.99 C

4 0

3 years ago

Solute particles can be:____

Reptile [31]

Solute particles can be atoms, ions or molecules.

Explanation:

Solute is the material which has to be mixed in the solvent to prepare a solution. So the concentration of solute should be less than the solvent. Also the solute and solvent should be of same nature other they will not dissolve with each other. The solute can be made up of atoms, ions or molecules depending upon the solvent. If the solvent concentration is in moles, then the solute concentration can be taken as atoms, ions or molecules. Also the saturation point plays a main role in deciding the kind of particles taken for the solute.

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3 years ago

Give the formula

madreJ [45]

Hello!!! I would like to give you my answer!!!
<span>%error= (-890kJ-0.07kJ)/(-890kJ)x100 = 110.02%
I hope this helps!!! :-D</span>

8 0

3 years ago

Why does nitrogen have stronger attractive forces between its nucleus and valence electrons than phosphorus?

timofeeve [1]

Answer:

Nitrogen atom is small in size than phosphorus thus the bond between its nucleus and valence electrons are stronger than phosphorus.

Hope it helps...｡◕‿◕｡

3 0

3 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

6 0

3 years ago

Why do we need to calibrate equipment​

Why do we need to calibrate equipment