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3 years ago

Match the following names of glassware with what you would use them for.

Chemistry

1 answer:

Valentin [98]3 years ago

3 0

Answer:

A) Graduated pipette – Glassware used to accurately transfer small volumes.

B) Volumetric pipette – Glassware used to accurately transfer a small, single volume.

C) Beaker – Glassware best used when greater access to the contents is needed.

D) Buret – Glassware used to deliver a volume not known in advance.

E) Erlenmeyer flask – Glassware used to prevent splashing or evaporation.

F) Volumetric flask – Glassware used to make accurate solutions.

Explanation:

Graduated pipette – Glassware used to accurately transfer small volumes.

A graduated pipette is a pipette, which has a scale that shows its volume marked along the tube. It is used to transfer small volumes accurately.

Volumetric pipette – Glassware used to accurately transfer a small, single volume.

A volumetric pipette is a pipette, which has a ring like marking that is its calibrated volume. So it is used to transfer a single and small volume only. This pipette is used in volumetric analysis.

Beaker – Glassware best used when greater access to the contents is needed.

Beaker is the most widely used glassware in the laboratory. They are used to transfer large volume with less accuracy. They are of different sizes depends on the size of volumes ranging from 10 mL to 1000 mL.

Buret – Glassware used to deliver a volume not known in advance.

Buret is the most important glassware in the quantitative analysis. It has a glass tube with scale which measures the volume and a stopcock at one end from which the solvent is dispersed. It is used to measure the volume of the liquid during the titration in the quantitative analysis.

Erlenmeyer flask – Glassware used to prevent splashing or evaporation.

The most common names of Erlenmeyer flask are conical flask and titration flask. This flask has flat bottom, conical body and cylindrical neck which prevent splashing and evaporation. This flask is used in the titration process in the quantitative analysis. The solvent from the buret is delivered into the conical flask during the titration process.

Volumetric flask – Glassware used to make accurate solutions.

The volumetric flask is also an important glassware in the analytical laboratory. It is used to prepare standard solutions. It is a flask which has a ring like marking that is its calibrated volume. The mentioned volume of volumetric flask is calibrated to have accurate volume.

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3 years ago

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tatuchka [14]

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3 years ago

A 0.223 mole sample of gas is held at 33.0 C and 2.00 atm, What's the volume of the gas? R = 0.0821 L atm / mol K answer soon il

Ghella [55]

Answer:

The volume of the gas is 2.80 L.

Explanation:

An ideal gas is a theoretical gas that is considered to be made up of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The Pressure (P) of a gas on the walls of the container that contains it, the Volume (V) it occupies, the Temperature (T) at which it is located and the amount of substance it contains (number of moles, n) are related from the equation known as Equation of State of Ideal Gases:

P*V = n*R*T

where R is the constant of ideal gases.

In this case:

P= 2 atm
V= ?
n=0.223 moles
R= 0.0821 $\frac{L*atm}{mol*K}$
T=33 °C= 306 °K (being O°C= 273°K)

Replacing:

2 atm* V= 0.223 moles*0.0821 $\frac{L*atm}{mol*K}$ * 306 K

Solving:

$V=\frac{0.223 moles*0.0821\frac{L*atm}{mol*K} * 306 K}{2 atm} \\$

V= 2.80 L

The volume of the gas is 2.80 L.

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3 years ago

Consider a general reaction

choli [55]

Answer:

a) K = 5.3175

b) ΔG = 3.2694

Explanation:

a) ΔG° = - RT Ln K

∴ T = 25°C ≅ 298 K

∴ R = 8.314 E-3 KJ/K.mol

∴ ΔG° = - 4.140 KJ/mol

⇒ Ln K = - ( ΔG° ) / RT

⇒ Ln K = - ( -4.140 KJ/mol ) / (( 8.314 E-3 KJ/K.mol )( 298 K ))

⇒ Ln K = 1.671

⇒ K = 5.3175

b) A → B

∴ T = 37°C = 310 K

∴ [A] = 1.6 M

∴ [B] = 0.45 M

∴ K = [B] / [A]

⇒ K = (0.45 M)/(1.6 M)

⇒ K = 0.28125

⇒ Ln K = - 1.2685

∴ ΔG = - RT Ln K

⇒ ΔG = - ( 8.314 E-3 KJ/K.mol )( 310 K )( - 1.2685 )

⇒ ΔG = 3.2694

7 0

3 years ago

112 g of aluminum carbide react with 174 g water to produce methane and aluminum hydroxide in the reaction shown below.

dolphi86 [110]

Answer: 4.999 moles of excess reactant will be left over.

Explanation:

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

$\text{Moles of aluminium carbide}=\frac{112g}{143.96g/mol}=0.778mol$

For the given chemical reaction:

$2Al_4C_3(s)+12H_2O(l)\rightarrow 3CH_4(g)+4Al(OH)_3(s)$

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = $\frac{12}{2}\times 0.778=4.668 mol$ of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

$\text{Moles of water}=\frac{174g}{18g/mol}=9.667mol$

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

8 0

2 years ago