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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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En el laboratorio, un vaso precipitado vacío tiene una masa de 0,50 kg. Este vaso se llena con 300 cm3 de una solución saturada
faust18 [17]

Answer:

Density = 0.0007667 Kg/cm³

Explanation:

Given the following data;

Mass of empty beaker = 0.50 kg

Volume of solution = 300 cm³

Mass of glass+solution = 730 g

To find the density of the solution in the glass;

First of all, we would convert the value in grams to kilograms.

Conversion:

1000 grams = 1 kg

730 = X kg

Cross-multiplying, we have;

X = 730/1000 = 0.73 kg

Next, we would determine the mass of the solution.

Mass of solution = Mass of glass+solution - Mass of empty beaker

Mass of solution = 0.73 - 0.50

Mass of solution = 0.23 kg

Lastly, we would solve for the density of the solution;

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the formula;

Density = \frac {mass}{volume}

Substituting into the formula, we have;

Density = \frac {0.23}{300}

Density = 0.0007667 Kg/cm³

4 0
3 years ago
what noble gas has the same electron configuration as each of the ions in the following compounds? a. Cesium sulfide b. strontiu
saveliy_v [14]

Answer:

A. Cesium sulfide

Explanation:

looked it up lol

7 0
3 years ago
Ngl I have no clue what I’m doing can someone help me pls this is due liek in 15 mins
Ad libitum [116K]
1. Calcium & Nitrogen: Ca3N2
2. Aluminum & Chlorine: AlCl3
3. Aluminum & Nitrogen: AlN
4. Potassium & Bromine: KBr
5. Magnesium & Oxygen: MgO
6. Sodium & Sulfur: Na2S
5 0
3 years ago
A chemist must dilute 34.3mL of 1.72mM aqueous calcium sulfate solution until the concentration falls to 1.00mM. He'll do this b
EastWind [94]

Answer:

58.9mL

Explanation:

Given parameters:

Initial volume  = 34.3mL    = 0.0343dm³

Initial concentration  = 1.72mM   = 1.72 x 10⁻³moldm⁻³

Final concentration  = 1.00mM = 1 x 10⁻³ moldm⁻³

Unknown:

Final volume  =?

Solution:

Often times, the concentration of a standard solution may have to be diluted to a lower one by adding distilled water. To find the find the final volume, we must recognize that the number of moles of the substance in initial and final solutions are the same.

   Therefore;

              C₁V₁  =  C₂V₂

where C and V are concentration and 1 and 2 are initial and final states.

        now input the variables;

                      1.72 x 10⁻³ x  0.0343 = 1 x 10⁻³  x V₂

                        V₂ = 0.0589dm³ = 58.9mL

         

4 0
3 years ago
How do you use the changes in the phase of water to keep you cool?
sineoko [7]

Answer:

by freezing

Explanation:

8 0
3 years ago
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