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.
Answer:
D
Explanation:
it doesn't matter how many protons k has all the answers have electron with the positive charge when is negative the question is trying to distract you D is the only one that would use the electrons sign correctly
1) Magnesium Chloride
2) Sodium Bromide
3) Magnesium Oxide
4) Nickel (III) Fluoride
5) Aluminum Chloride
6) <span>Rubidium Sulfide
7) Gallium Nitride
8) Calcium Sulfide
9) </span><span>Lead (IV) Oxide
10) </span><span>Cobalt (II) Oxide
</span>11) B<span>eryllium Sulfide
12) </span><span>Cesium Nitride</span>
Answer: 72L of 30% and 128L of 80%
You can determine the weight of the acid by multiplying the concentration with the volume. Let say v1 is the volume of 30% solution needed and v2 is the volume of 80% solution.
The weight of acid from the used solution should be equal to the product. You can get this equation
final solution= solution1 + solution2
200l * 62%= v1 * 30% + v2*80%
124L= 0.3v1 + 0.8v2
124L- 0.3v1= 0.8v2
v2=155L- 0.375v1
The total volume of both should be 200l. If you use the previous equation, you can calculate:
v1+v2=200L
v1+ (155L- 0.375v1)= 200L
0.625v1= 200L - 155L
v1= 45/ 0.625= 72L
v1+v2=200L
v2= 200L- 72L= 128L
