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Gennadij [26K]
3 years ago
15

When heating a sample to extreme heat over a flame, you will often use a crucible. Before starting the heating, you should locat

e the correct labware for the experiment, including the equipment to safely handle and support the crucible. What equipment should you have ready to start your crucible experiment
Chemistry
1 answer:
Ghella [55]3 years ago
6 0

Answer:

The equipments you should have ready to start the crucible experiment includes: safety goggles, crucible with lid, crucible tong, ring support with clay triangle, Bunsen burner and heat resistant tile.

Explanation:

Crucible is an equipment in the laboratory which is suitable for heating a sample to extreme heat over a flame, Modern laboratory crucible are made up of graphite- based composite materials for achievement of higher performance. Because extreme heat is involved, you should locate the correct labware for the experiment, including the equipment to safely handle and support the crucible. These equipments includes:

--> Safety goggles: Because you will work with chemical it is advisable to use a safety goggles which protects the eyes from dangerous floating chemical aerosol.

--> crucible with lid: This is the main apparatus with the lid (cover) which is used to cover the crucible to prevent spilling of the boiling chemical.

--> Crucible tong: These are scissors like tools used to grasp hot crucible.

--> Ring support with clay triangle: the clay triangle is used to hold crucible when they are being heated. They usually sit on a ring stand.

--> Bunsen burner: Produces a single open gas flame which can be used for heating.

With the safety equipments listed above, you can carry out experiment using the crucible. These equipments helps minimise laboratory hazard that may occur should Incase it's not available.

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It releases nutrients which causes rapid growth of planktons which leads to oxygen deficiency leading to death of aquatic animals,a phenomenon known as eutrophication.

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Given the data in the table, what is the average atomic mass of chlorine?
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Which equation shows how to calculate how many grams (g) of Mg(OH)2 would be produced from 4mol KOH? The balanced reaction is:
Simora [160]

Answer:

4 mol of KOH would produce 116.6 g of Mg(OH)₂

Explanation:

According to the following balanced equation:

  • MgCl₂+ 2 KOH -----> Mg(OH)₂ + 2 KCL

One can note that 2 mol of KOH react with MgCl₂ to produce 1 mol of Mg(OH)₂.

using cross multiplication  

2 mol of KOH → 1 mol of Mg(OH)₂.

4 mol of KOH → ?? mol of Mg(OH)₂.

no of moles of  Mg(OH)₂ = (1 mol* 4 mol) / 2 mol =2 mol

Now we can convert moles of  Mg(OH)₂ to grams using the formula

mass of Mg(OH)₂= (no. of moles * molar mass) = (2 mol * 58.3g/mol) = 116.6 g

  • So, 4 mol of KOH would produce 116.6 g of Mg(OH)₂.
6 0
4 years ago
A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure
Lesechka [4]

Answer:

the final volume of the gas is V_2 = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure P_{ext}:

P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}

P_{ext} = 1.03 \ atm

The workdone W = P_{ext}V

The change in volume ΔV= \dfrac{W}{P_{ext}}

ΔV = \dfrac{130.1 \ J  \times \dfrac{1 \ L  \ atm}{ 101.325 \ J}  }{1.03 \ atm }

ΔV = \dfrac{1.28398717 }{1.03  }

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial  volume = 61.5 mL

The change in volume V is \Delta V = V_2 -V_1

-  V_2= -  \Delta V  -V_1

multiply through by (-), we have:

V_2=   \Delta V+V_1

V_2 =  1250 mL + 61.5 mL

V_2 = 1311.5 mL

∴ the final volume of the gas is V_2 = 1311.5 mL

5 0
3 years ago
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