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Temka [501]
3 years ago
5

1. Describe the general procedure on how to wash glassware used for organic lab experiments. Also, include what the common solve

nt used to wash organic glassware
Chemistry
1 answer:
Gemiola [76]3 years ago
5 0

Explanation:

As a number of glassware are used in lab experiments so it is necessary that they should be cleaned properly after and before the experiment.

This can be done as follows.

1). At first acetone is used to rinse the glassware. If water soluble contents present in the glassware then use deionized water after rinsing it with acetone and if ethanol soluble components are present there then rinse with ethanol followed by rinses with deionized water.

2). When glassware are dirty in such a manner that they cannot be washed immediately then soak them in water for a certain period of time. This can help in removing the dirt or chemicals easily from the glassware.

3). The laboratory glassware can be easily washed with detergents or products like lab wash, alconox etc. Once these glassware are thoroughly cleaned then rinse them 3 times with deionized water.

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Help plz I will give u BRAINLIST
Agata [3.3K]

CS2 + 3O2 = CO2 + 2SO2

1 mole of CS2 gives 1 mole of CO2

12 + 2(32) = 76g of CS2  yields 44 g of CO2

Theoretically 1 g of CS2 yields 44/76 g CO2

Therefore 50 g CS2 should yield    50*44 / 76 = 28.95 g

So % yield = 103.6 %  ( which is not possible  because you can't create matter from nothing).

The 30g cannot be right . This is experimental err.

6 0
3 years ago
Cual es el numero de oxidacion de la respiración​
Vinil7 [7]
The answer is 0 if im right
3 0
3 years ago
Convert 3.30 g of copper (II) hydroxide Cu(OH)2 to molecules.
ratelena [41]

Answer:

0.18× 10²³ molecules

Explanation:

Given data:

Mass of copper hydroxide = 3.30 g

Number of molecules = ?

Solution:

Number of moles = mass/molar mass

Number of moles = 3.30 g/97.56 g/mol

Number of moles = 0.03 mol

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

0.03 mol × 6.022 × 10²³ molecules / 1mol

0.18× 10²³ molecules

6 0
2 years ago
IS THISS RIGHTTTTT???
shtirl [24]

Answer: No

Explanation: For it to be a divergent boundary, the arrows would have to be pointing in opposite directions. (one points left, one points right).

3 0
3 years ago
Read 2 more answers
A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

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

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0
3 years ago
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