Atomic Number of Lithium is 3, so it has 3 electrons in its neutral state. Also, Li₂ will have 6 electrons. But the chemical formula we are given has a negative charge on it (i.e Li₂⁻) so there is an additional electron (RED) present on this compound. So, the total number of electrons are 7. The
MOT diagram for this compound is shown below. According to diagram we are having 4 electrons in Bonding Molecular Orbitals (
BMO) and 3 electrons in Anti-Bonding Molecular Orbitals (
ABMO). Bond Order is calculated as,
Bond Order = (# of e⁻s in BMO - # of e⁻s in ABMO) ÷ 2
Bond Order = (4 - 3) ÷ 2
Bond Order = 1 ÷ 2
Or,
Bond Order = 1/2Or,
Bond Order = 0.5
Answer is: an oxybromate compound is KBrO₄ (x = 4).
ω(Br) = 43.66% ÷ 100%.
ω(Br) = 0.4366; mass percentage of bromine.
If we take 100 grams of compound:
m(Br) = ω(Br) · 100 g.
m(Br) = 0.4366 · 100 g.
m(Br) = 43.66 g; mass of bromine.
n(Br) = m(Br) ÷ M(Br).
n(Br) = 43.66 g ÷ 79.9 g/mol,
n(Br) = 0.55 mol; amoun of bromine.
From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).
m(K) = 0.55 mol · 39.1 g/mol.
m(K) = 21.365 g; mass of potassium in the compound.
m(O) = 100 g - 21.365 g - 43.66 g.
m(O) =34.97 g; mass of oxygen.
n(O) = 34.97 g ÷ 16 g/mol.
n(O) = 2.185 mol.
n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.
n(K) : n(Br) : n(O) = 1 : 1 : 4.
Answer:
The maximum pressure is 612.2 Pa
Explanation:
The pressure of the ice (P1) = 624 Pa
The temperature of the ice = 273.16 K
The maximum temperature the specimen = - 5 oC
= -5 + 273 = 268K
The maximum Pressure the freeze drying can be will be (P2) = ?
Using Pressure law, which shows the relationship between pressure and temperature.
P1 / T1 = P2 / T2
P2 T1 = P1 T2
P2 = P1 T2 / T1
P2 = 624 × 268 / 273.16
P2 = 612.2 Pa
The maximum pressure at which drying can be carried out is 612.2 Pa
Check the attached document more explanation. jjjjggggg
Answer:
C
Explanation:
Wind is geological therefore it is geological weathering
Explanation:
speed = distance/time
= 23.7/54 m/s
= 0.44 m/s
speed of a dog running through a field = 0.44 m/s
