Answer : The half life of 28-Mg in hours is, 6.94
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = time passed by the sample = 48.0 hr
a = initial amount of the reactant disintegrate = 53500
a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520
Now put all the given values in above equation, we get
Now we have to calculate the half-life.
Therefore, the half life of 28-Mg in hours is, 6.94
Oxygen, fluorine and iodine are diatomic elements. Flourine is more reactive than the other two because it is the closest away to filling its outer layer of electrons and becoming stable like a noble gas.
Well, an element is a substance that cannot be decomposed into a simpler substance by chemical change, so element is the answer you are probably looking for! Hope this helps!
When pcl5 solidifies it forms pcl4 cations and pcl6–anions. according to valence bond theory, hybrid orbitals that are used by phosphorus in the pcl4 cations are one orbital of s and three orbital of p as it is sp³hydridised.
<h3>What is sp³ hybridization?</h3>
Hybridization is a process or system which specifies the shape and geometry of any element or molecule with bond angles too.
The pcl4 cation is sp³ hybridized because of the phosphorus behave as a central atom here and the 4 chloride molecules are attached with the p- orbitals to the phosphorus molecule.
Therefore, pcl4 cation is sp³ hybridized.
Learn more about sp³ hybridization, here :
brainly.com/question/13062274
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Combustion reaction occurs when organic compound reacts with oxygen to form CO₂, H₂O and energy
C(s) + O₂(g) → CO₂(g) is Synthesis reaction not combustion
2 H₂(g) + O₂(g) → 2 H₂O(g) also Synthesis reaction and not combustion
C₃H₈(g) + 5 O₂ → 3 CO₂(g) + 4 H₂O(g) is considered as Combustion reaction
2 C₃H₇OH(l) + 9 O₂(g) → 6 CO₂(g) + 8 H₂O(g) Combustion reaction
