Explanation:
1.
Given parameters:
Frequency of the radiation = 8.4 x 10¹⁴Hz
Unknown:
Energy of the wave = ?
Solution:
The energy of a wave is given by the expression below;
E = hf
E is the energy
h is the Planck's constant = 6.63 x 10⁻³⁴m²kg/s
f is the frequency
Now insert the parameters and solve;
E = 6.63 x 10⁻³⁴m²kg/s x 8.4 x 10¹⁴Hz
E = 5.57 x 10¹ x 10⁻²⁰J
E = 5.57 x 10⁻¹⁹J
2.
Given parameters:
Wavelength = 2.13 x 10⁻¹³m
Unknown:
Frequency of the wave = ?
Solution:
The frequency of a wave can be determined using the expression;
C = f∧
C is the speed of light = 3 x 10⁸m/s
f is the frequency
∧ is the wavelength
f =
=
= 1.41 x 10²¹hz
Answer:
The answer to the question is
The specific heat capacity of the alloy = 1.77 J/(g·°C)
Explanation:
To solve this, we list out the given variables thus
Mass of alloy = 45 g
Initial temperature of the alloy = 25 °C
Final temperature of the alloy = 37 °C
Heat absorbed by the alloy = 956 J
Thus we have
ΔH = m·c·(T₂ - T₁) where ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C , T₂ = Final temperature of the alloy = 37 °C and m = mass of the alloy = 45 g
∴ 956 J = 45 × C × (37 - 25) = 540 g·°C×c or
c = 956 J/(540 g·°C) = 1.77 J/(g·°C)
The specific heat capacity of the alloy is 1.77 J/(g·°C)
A synthesis would be A+B-->AB its combining, decomposition is AB-->A+B, its the breaking up of a substance.
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.