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2 years ago

Which of the following is the best piece of laboratory glassware for preparing 500.0 mL of an aqueous solution of a solid?

Chemistry

2 answers:

Leni [432]2 years ago

8 0

Answer:

The correct answer is option A.

Explanation:

Volumetric flask : A glass ware with round lower body with flat bottom and with thin cylindrical neck along with mark which indicates the specific volume filled to that mark.It is used in preparation of standard solution of compound with desired concentration with fixed volume.

Erlenmeyer flask : is a flask with conical shape with flat bottom used in titration experiments to carry out reaction with fixed volume of solution.

Test tube : Small cylindrical tube with rounded bottom used to observe reaction in between reactant taken in small amount.

Graduated beaker : Laboratory glassware used measure larger volumes of solution or to mix or stir solutions and liquids.

Graduated cylinder : Laboratory thin cylindrical glassware with accurate marking of volume used to measure an accurate volume of solutions or liquids required in an experiment.

<em><u>Volumetric flask</u></em> is the best piece of laboratory glassware for preparing 500.0 mL of an aqueous solution of a solid

GalinKa [24]2 years ago

8 0

Answer:

its A

Explanation:

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A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final

alekssr [168]

Answer:

208.7°C was the initial temperature of the limestone.

Explanation:

Heat lost by limestone will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of limestone = $m_1=62.6 g$

Specific heat capacity of limestone = $c_1=0.921 J/g^oC$

Initial temperature of the limestone = $T_1=?$

Final temperature = $T_2$ =T = 51.9°C

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2=75.0 g$

Specific heat capacity of water= $c_2=4.186 J/g^oC$

Initial temperature of the water = $T_3=23.1^oC$

Final temperature of water = $T_2$ =T = 51.9°C

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(62.6 g\times 0.921 J/g^oC\times (51.9^oC-T_1))=75.0 g\times 4.186 J/g^oC\times (51.9^oC-23.1^oC)$

$T_1=208.7^oC$

208.7°C was the initial temperature of the limestone.

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2 years ago

When sodium chloride is dissolved in water, the resulting solution is classified as a?

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A homogeneous mixture is formed when it is dissolved in water

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2 years ago

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kramer

Answer:

1..... nucleus

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4........Neutrons

5..........electron

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2 years ago

Balance the equation AlCl3 + H2SO4 yields Al2(SO4)3 +HCl

Natalija [7]

Answer:

2AlCl3 + 3H2SO4 → Al2(SO4)3 + 6HCl

Explanation:

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2 years ago

A sample of oxygen gas is compressed from 30.6 L to 1.8 L at constant temperature pressure of 1.8 atm. Calculate the amount of e

ad-work [718]

Answer:

the change in the internal energy of the system is 3,752.67 J

Explanation:

Given;

initial volume of the gas, V₁ = 30.6 L

final volume of the gas, V₂ = 1.8 L

constant pressure of the gas, P = 1.8 atm

Energy released by the system, Q = 1.5 kJ = 1,500 J

Apply pressure-volume work equation, to determine the work done on the gas;

w = -PΔV

w = -P(V₂ - V₁)

w = - 1.8 atm(1.8 L - 30.6 L)

w = 51.84 L.atm

w = 51.84 L.atm x 101.325 J/L.atm

w = 5,252.67 J

The change in the internal energy of the system is calculated as;

ΔU = Q + w

Since the heat is given out, Q = - 1,500 J

ΔU = -1,500 J + 5,252.67 J

ΔU = 3,752.67 J

Therefore, the change in the internal energy of the system is 3,752.67 J

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2 years ago