1. 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂
2. CH₄ + 2O₂ → CO₂ + 2H₂O
3. Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag
4. MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
5. Pb(NO₃)₂ + 2NaCl → PbCl₂ + 2NaNO₃
1)

2)
CuSO_4+Cu_2Cl_2\neq>

Answer:
a solute is a substance that is dissolved in the solvent (liquid)
a solvent is a substance that dissolves a solute (solid)
a solution is the mixture of both the solute and solvent forming uniform mixture
The question is incomplete, here is the complete question:
A VHF television station assigned to channel transmits its signal using radio waves with a frequency of 66.0 MHz. Calculate the wavelength of the radio waves. Round your answer to 3 significant digits.
Answer: The wavelength of the radio waves is 4.54 m
<u>Explanation:</u>
To calculate the wavelength of light, we use the equation:

where,
= wavelength of the light
c = speed of light = 
= frequency of light = 
Putting the values in above equation, we get:

Hence, the wavelength of the radio waves is 4.54 m
Answer:
Work done by the system = 4545 J
Explanation:
The expression for the calculation of work done is shown below as:
Where, P is the pressure
is the change in volume
From the question,
= 45 - 15 L = 30 L
P = 1.5 atm
Also, 1 atmL = 101 J
So,
(negative sign implies work is done by the system)
<u>Work done by the system = 4545 J</u>
The main information we have to use here is the density of gold. From literature, the density of gold at room temperature is 19.32 g/cm³. To determine the mass, let's calculate the volume first. A wire is in the shape of a cylinder. Thus, the volume would be
V = πd²h/4
V = π(0.175 cm)²(1×10⁵ cm)/4
V = 2,405.28 cm³
Density = mass/volume
19.32 g/cm³ = Mass/2,405.28 cm³
Mass = 46,470 g gold wire