Answer:
Na2SO4
Explanation:
In the crisscross method, we can write the correct formula for an ionic compound by crossing over the numerical value of each of the ion charges to now serve as the subscript of the oppositely charged ion. Subsequently, the signs of the charges are dropped and the correct formula of the ionic substance is obtained.
For sodium sulphate;
Na ^+ SO4^2-
By crisscrossing and dropping the signs to obtain the subscripts we now have;
Na2SO4
The balanced nuclear equation for the reaction is
<h3>²³⁵₉₂U + ¹₀n —> ¹⁵⁵₆₂Sm + ⁷⁸₃₀Zn + 3(¹₀n)</h3>
From the question given above, we were told that:
<u>A fast-moving neutron strikes a ²³⁵U nucleus. The nucleus shatters producing ¹⁵⁵Sm, ⁷⁸Zn, and three neutrons.</u>
The nuclear equation can be written as follow:
Neutron => ¹₀n
Uranium => ²³⁵₉₂U
Samarium => ¹⁵⁵₆₂Sm
Zinc => ⁷⁸₃₀Zn
Uranium + neutron —> Samarium + Zinc + 3 moles of neutron
<h3>²³⁵₉₂U + ¹₀n —> ¹⁵⁵₆₂Sm + ⁷⁸₃₀Zn + 3(¹₀n)</h3>
The nuclear equation above is balanced.
Learn more: brainly.com/question/9943790
Entropy Change is the measure of the randomness of the thermodynamic system. The entropy change for the reaction is -49.3 J/K.
<h3>What is the change in entropy?</h3>
Entropy change is the ratio of the heat transfer of the system to the absolute temperature of the system.
The entropy of nickel = 182.1 J/mol. K
The entropy of oxygen = 205.0 J/mol. K
The entropy of nickel oxide = 37.99 J/mol. K
Entropy change is calculated as:

Therefore, -49.3 J/K is the entropy change.
Learn more about entropy change here:
brainly.com/question/1301642
#SPJ4
Your first step should be to analyse the compound. For example, if the compound is carbon, you know it always has a valence of four, so, if it has a formula C2H4 (ethylene) it obviously has a double bond. There are difficulties here because benzene C6H6 can be considered to have 6 1.5 C-C bonds, being aromatic.
A second step is to look at its structure. Double bonds are traditionally shorter than single bonds; triple bonds shorter still. Covalent bonds do have typical lengths, nevertheless you can still have problems.
<span>A third step is to consider reactivity. For example, if you have a C=C double bond, you can add, say, bromine to it Thus C2H4 gives C2H4Br2, and by adding two bromine atoms you know you have one double bond. Again, benzene becomes an awkward molecule, but because of this, you know benzene does not have double bonds in the traditional sense</span>