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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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No

Explanation:

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An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at
lana [24]

Answer:

ΔU = −55.45 kJ

Explanation:

From first law of thermodynamics in chemistry, we have;

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the net heat transfer

W is the net work done

We are given;

Q = 74.6 kJ

But Q will be negative since heat is released

Thus;

ΔU = -74.6 kJ + W

We are given;

Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²

Volume before reaction; Vi = 8.2 L = 0.0082 m³

Volume after reaction; V_f = 2.8 L = 0.0028 m³

Now,

W = -P(V_f - V_i)

W = - 3546375(0.0028 - 0.0082)

W = 19.15 KJ

Thus;

ΔU = Q + W

ΔU = -74.6 kJ + 19.15 KJ =

ΔU = −55.45 kJ

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Calculate the work associated with the compression of a gas from 121.0 L to 80.0 L at a constant pressure of 16.7 atm.
Mkey [24]

Answer:

6.94 × 10⁴ J

Explanation:

We can calculate the work (W) associated with the compression of a gas at constant pressure using the following expression:

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where,

P is the external pressure

ΔV is the change in the volume

Knowing that 1 atm.L = 101.325 J, the work of compression is:

W=-P\times \Delta V=-16.7atm\times(80.0L-121.0L)\times\frac{101.325J}{1atm.L} =6.94\times10^{4} J

The positive sign of the work means that the surroundings do work on the system.

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3 years ago
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