The hybrid orbital of this molecule is 
. Hence, option C is correct.
<h3>What is hybridisation?</h3>
Hybridization is defined as the concept of mixing two atomic orbitals to give rise to a new type of hybridized orbitals.
In this compound, a hybrid orbital makes I-O bonds. Due to hybridization iodate should have tetrahedral geometry but because of the presence of lone pair of electrons the shape of 
the ion is pyramidal.
The hybrid orbital of this molecule is . Hence, option C is correct.
Learn more about hybridisation here:
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We can use the formula P=IV to calculate the current, where “P” is power (measured in watts), “I” is current (measured in Amps), and “V” is voltage. Simply plug and solve:
P = IV
(3.5 Watts) = I(120 volts)
I = 0.0292 Amps
The current flowing through the bulb is approximately 0.0292 Amps.
Hope this helps!
To solve this, let's assume ideal gas behavior.
PV=nRT
Let's solve for n. Convert units to SI units first.
Pressure = 833 torr(101325 Pa/760 torr) = 111,057.53 Pa
Volume = 250 mL(1 L/1000 mL)(1 m³/1000 L) = 2.5×10⁻⁴ m³
Temperature = 42.4 + 273 = 315.4 K
n = (8,314 J/mol·K)(315.4 K)/(111057.53 Pa)(2.5×10⁻⁴ m³)
n = 94.45 mol
The molar mass of ammonia is 17.031 g/mol.
Mass = 94.45*17.031 = <em>1,608.51 g ammonia</em>
Lanthanide Lanthanoid, also called Lanthanide
