We can calculate how long the decay by using the half-life equation. It is expressed as:
A = Ao e^-kt
<span>where A is the amount left at t years, Ao is the initial concentration, and k is a constant.
</span><span>From the half-life data, we can calculate for k.
</span>
1/2(Ao) = Ao e^-k(30)
<span>k = 0.023
</span>
0.04Ao = Ao e^0.023(t)
<span>t = 140 sec</span>
Answer: Ions may be defined as the element that contains either positive or negative charge over them. Two types of ions are cations and anions. The outermost electrons are involved in the formation of ions.
The atomic number of sulfur is 16. Its outermost electronic configuration is K=2, L= 8, M= 6. The sulfur requres two more electrons to complete its orbit and accquire -2 charge.
Explanation:
This problem is requiring the empirical formula for CaCO₃, which is its molecular formula, and turns out to be equal, this is A. CaCO3 according to the following:
<h3>Empirical formulas:</h3><h3 />
In chemistry, molecular formulas show both the actual type and number of atoms in a chemical compound, based on the elements across the periodic table and the subscripts standing for the number of atoms in the compound.
However, the empirical formula is a reduced expression of the molecular one, which shows the minimum number of atoms in a compound after simplifying to the smallest whole numbers.
In such a way, since the given compound is CaCO₃ and both Ca and C have a one as their subscript, it is not possible to simplify any further and therefore the empirical formula equals the molecular one this time, making the answer to be A. CaCO3.
Learn more about empirical formulas: brainly.com/question/1247523
Answer:
3 P atoms
Explanation:
Al₂P₃ => contains 2 Aluminum ions (2Al⁺³) and 3 Phosphide ions (3P⁻³) ... The ions (charged particles) are from atoms that have lost or gained electrons during the bonding process. So, Al₂P₃ => P⁻³ ions from 3 P atoms.
Explanation:
When conducting a melting point experiment, if we were to heat a sample quickly. Large amount heat is provided instantly which would melt the crystals in the tube very quickly, even before the temperature of the thermometer reaches to that level. So the observes melting point would be much lower than the actual melting point when sample is heated slowly.
