Answer:

Explanation:
Given that:
The Half-life of
=
is less than that of 
Although we are not given any value about the present weight of
.
So, consider the present weight in the percentage of
to be y%
Then, the time elapsed to get the present weight of
= 
Therefore;

here;
= Number of radioactive atoms relating to the weight of y of 
Thus:

--- (1)
However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of
to be = 
Then:
---- (2)
here;
= Number of radioactive atoms of
relating to 3.0 a/o weight
Now, equating equation (1) and (2) together, we have:

replacing the half-life of
=
( since
)
∴

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o
Thus, The time elapsed is 
I think It’s 55 but that’s just me
Answer:
See below
Explanation:
<u> Name </u> <u>Formula </u> <u> Major species </u> <u> </u>
Zinc iodide ZnI₂ H₂O(ℓ), I⁻(aq), Zn²⁺(aq),
Nitrogen(I) oxide N₂O H₂O(ℓ), N₂O(aq)
Sodium nitrite NaNO₂ H₂O(ℓ), Na⁺(aq), NO₂⁻(aq)
Glucose C₆H₁₂O₆ H₂O(ℓ), C₆H₁₂O₆(aq)
Nickel(II) iodide NiI₂ H₂O(ℓ), I⁻(aq), Ni²⁺(aq)
- Glucose and nitrogen(I) oxide are covalent compounds. They do not dissociate in solution.
- The compounds containing metals are ionic. They produce ions in solution.
- ZnI₂ and NiI₂ produce twice as many iodide ions as metal ions.
<u>Answer:</u> Increasing temperature
<u>Explanation:</u>
The Principle of Le Chatelier states that <u>if a system in equilibrium is subjected to a change of conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.
</u>
The variation of one or several of the following factors can alter the equilibrium condition in a chemical reaction:
- Temperature
- The pressure
- The volume
- The concentration of reactants or products
In the case of the reaction in the question, <u>the change that moves the balance to the left will be the one that moves it towards the reagents</u>, that is, that favors the production of reagents instead of products.
-
Decreasing the concentration of SO3 and increasing the concentration of SO2 <u>will favor the production of SO3</u>, which is the product of the reaction.
- Decreasing the volume increases the pressure of the system and the balance will move to where there is less number of moles. In the case of the reaction in question, we have 3 moles of molecules in the reactants (1 mole of O2 + 2 moles of SO2) while in the products there are 2 moles of SO3 only, therefore, <u>decreasing the volume will displace the balance to the right</u>, which corresponds to the sense in which there is less number of moles.
The reaction of the question is an exothermic since ΔH <0, therefore in the reaction heat is produced and it can be written in the following way,
2SO2(g) + O2(g) ⇌ 2SO3(g) + heat
- So, if we increase the temperature we will be adding heat to the system, so the balance would move to the left to compensate for the excess heat in the system.