Long wavelengths and low frequencies
Answer:
Zero
Explanation:
FrBr is an ionic compound
.
Fr is in Group 1. Br is in Group 17.
The charges on the ions are +1 and -1, respectively.
The compound consists of Fr⁺Br⁻ ions.
However, there are equal numbers of + and - charges, so
The overall charge of the compound is zero.
The substance that can be classified as being soluble in water would be salt. The solution is D. Salt specifically is composed of ions and the ionic compound dissociates within water.
Answer:
Here's what I get
Explanation:
1. Nickel sulfate
base + acid ⟶ salt + water
NiSO₄ is a salt of the base Ni(OH)₂ and the acid sulfuric acid.
Hydroxides of transition metals are insoluble; most sulfates are soluble.

2. Carbonate + acid
Most carbonates are insoluble.
They react with acids to form carbonic acid (H₂CO₃), which decomposes into water and carbon dioxide.

Answer:

Explanation:
1. Calculate the decay constant
The integrated rate law for radioactive decay is 1

where
A₀ and A_t are the counts at t = 0 and t
k is the radioactive decay constant

2. Calculate the half-life

The half-life for decay is
.