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Gennadij [26K]
2 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]2 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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Why is copper used in the manufacture of computers?
Stells [14]

IBM and others are using copper instead of aluminum in the most powerful computer chips they manufacture. Because of copper's superior electrical conductivity, this technology enables conductor channel lengths and widths to be significantly reduced.

3 0
2 years ago
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A chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water. What is the concentration of the final solution? A
anygoal [31]

this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.

c1v1 = c2v2

where c1 is concentration and v1 is volume of the concentrated solution

and c2 is concentration and v2 is volume of the diluted solution to be prepared

50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL

substituting these values in the formula

1.50 M x 50.0 mL = C x 250 mL

C = 0.300 M

concentration of the final solution is A) 0.300 M

4 0
3 years ago
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How many mols are present in 5.24 x 1024 atoms of Carbon tetra fluoride?
Free_Kalibri [48]

Answer: 5.44×10226.022140857(74)×1023⋅mol−1.

Explanation: So the answer is approx. 0.10⋅mol

3 0
2 years ago
The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in
astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}

For this, it is necessary to know the values in meters for any of these diameters:

\\ 1m = 10^{3}mm = 1e+3mm

\\ 1m = 10^{12}pm = 1e+12pm

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m

<h3>Diameter of a biscuit in meters</h3>

\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0
3 years ago
calculate the enthalpy change for converting 1.00 mol of liquid water at 100 C to water vapor at 145 degree celsius for water Δ
Svetllana [295]

Answer:

q1 = mCpΔT

    = 18.016g × 1.84J/g.K × (418.15-373.15)

     = 1491.72 J

q2 = n×ΔH vap

     =  1mol ×44.0kJ/mol

     = 44KJ

∴ qtotal = q1+ q2

             = 1.498kJ + 44.0kJ

             = 45.498KJ

Explanation: The heat flow can be separated into steps.all that is being observed at a constant pressure,the heat flow is equal to the enthalpy.

8 0
3 years ago
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