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3 years ago

How do you identify unknown chemicals.

Chemistry

1 answer:

QveST [7]3 years ago

7 0

Answer:

Preventing Unknown Chemicals

Label all chemical containers, including beakers, flasks, vials, and test tubes.

Immediately replace labels that have fallen off or that are deteriorated.

Label containers using chemical names. ...

Archived research samples are often stored in boxes containing hundreds of small vials.

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A solution of phosphoric acid was made by dissolving 10.0 g of H3PO4 in 100.0 mL of water. The resulting volume was 113 mL. Calc

rodikova [14]

Explanation:

Mass of solute = 10.0 g

mass of solvent(water) = m

Volume of solvent( water) = v = 100.0 mL

Density of water= d = $1 g/cm^3=1 g/mL$

$1 mL= 1 cm^3$

$m=d\times v=1.0 g/mL\times 100.0 = 100.0 g$

Mass of solution(M) = Mass of solute + mass of solvent

M = 10.0 g + 100.0 g = 110.0 g

Volume of the solution = V = 113 mL

Density of the solution = D

$D=\frac{M}{V}=\frac{110.0 g}{113 mL}=0.9734 g/mL$

The density of the solution is 0.9734 g/ml.

Moles of phosphoric acid = $n_1=\frac{10.0 g}{98 g/mol}=0.1020 mol$

Moles of water = $n_2=\frac{100.0 g}{18g/mol}=5.556 mol$

Mole fraction of phosphoric acid = $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1020 mol}{0.1020 mol+5.556 mol}$

$\chi_1=0.01803$

Mole fraction of water = $\chi_2$

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{5.556 mol}{0.1020 mol+5.556 mol}$

$\chi_2=0.9820$

$[Molarity]=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}$

Moles of phosphoric acid = 0.1020 mol

Volume of the solution = V = 113 mL = 0.113 L ( 1 mL = 0.001 L)

Molarity of the solution :

$=\frac{0.1020 mol}{0.113 L}=0.903 M$

$[Molality]=\frac{\text{Moles of solute}}{\text{Mass of solvent(kg)}}$

Moles of phosphoric acid = 0.1020 mol

Mass of solvent(water) = m =100.0 g = 0.100 kg ( 1 g = 0.001 kg)

Molality of the solution :

$=\frac{0.1020 mol}{0.100 kg}=1.02 mol/kg$

6 0

3 years ago

How many moles are in 12 grams of potassium chlorate, KClO3?

galina1969 [7]

It has one mole xxxxxxxx

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3 years ago

How much energy is needed to vaporize 75.0 g of diethyl ether (c4h10o) at its boiling point (34.6°c), given that δhvap of diethy

malfutka [58]

Answer: 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

Explanation:

First we have to calculate the moles of diethyl ether

$\text{Moles of diethyl ether}=\frac{\text{Mass of diethyl ether}}{\text{Molar mass of diethyl ether}}=\frac{75.0g}{74g/mole}=1.01moles$

As, 1 mole of diethyl ether require heat = 26.5 kJ

So, 1.01 moles of diethyl ether require heat = $\frac{26.5}{1}\times 1.01=26.8kJ$

Thus 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

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4 years ago

How molecules react to an increase in temperature.

Slav-nsk [51]

Answer:

An increase in temperature typically increases the rate of reaction. An increase in temperature will raise the average kinetic energy of the reactant molecules. Therefore, a greater proportion of molecules will have the minimum energy necessary for an effective collision (Figure. 17.5 “Temperature and Reaction Rate”).

Explanation:

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3 years ago

Which observation provided Albert Einstein the clue that he needed to explain the photoelectric effect? Light is made up of extr

Aliun [14]

The correct answer is

Energy of electrons depends on light’s frequency, not intensity.

As per photoelectric effect, if we incident a light on metal surface it will results into emission of electron from it

if we increase the number of photons the number of electrons will increase however if we increase the frequency the number of photons will not increase

While if we increase frequency the energy of electrons will increase as

Energy of photon = Work function of metal + kinetic energy of electrons

7 0

3 years ago

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