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:
F centripetal force (tension) = 275.9 N
Explanation:
Given data:
Mass = 1.50 kg
Radius = 0.520 m
Velocity of ball = 9.78 m/s
Tension = ?
Solution:
F centripetal force (tension) = m.v² / R
F centripetal force (tension) = 1.50 kg . (9.78 m/s)² / 0.520 m
F centripetal force (tension) = 1.50 kg . 95.65 m²/s² / 0.520 m
F centripetal force (tension) = 143.5 kg. m²/s² / 0.520 m
F centripetal force (tension) = 275.9 N
A is your answer.
On the periodic table the atomic number is the number of protons inside the nucleus.
Full question options;
(Fe, Pb, Mg, or Ca)
Answer:
Iron - Fe
Explanation:
We understand tht metals pretty much form bonds by losing their valence (outermost electrons). But this question specifically asks for metals that lose beyond their outermost electrons; next to outermost principal energy levels.
Pb, Mg, and Ca only lose their outermost electrons to form the following ions;
Pb2+, Mg2+, and Ca2+.
This is because their ions have achieved a stable octet configuration - the dreamland of atoms where they are satisfied and don't need to go into reactions again.
Iron on the other hand has the following electronic configurations;
Fe: [Ar]4s2 3d6
Fe2+: [Ar]4s0 3d6
Fe3+: [Ar]4s0 3d5
This means ion can lose both the ooutermost electrons (4s) and next to outermost principal energy levels (3d). So correct option is Iron.
PH=-lg[H⁺]
pH=-lg(2.8×10⁻⁹)=8.553
pOH=14-pH
pOH=14-8.553=5.447
[OH⁻]=10^(-pOH)
[OH⁻]=10⁻⁵·⁴⁴⁷=3.6×10⁻⁶
