Answer:
180 mg
Explanation:
For a first-order reaction, we can calculate the amount of aspirine (A) at a certain time (t) using the following expression.

where,
k: rate constant
A₀: initial amount
If we know the half-life (
) we can calculate the rate constant.

When t = 4 h and A₀ = 400 mg, A is:

<h3>
![\tt Kc=\dfrac{[CO_2]}{[C][O_2]}](https://tex.z-dn.net/?f=%5Ctt%20Kc%3D%5Cdfrac%7B%5BCO_2%5D%7D%7B%5BC%5D%5BO_2%5D%7D)
</h3><h3>Further explanation</h3>
Given
Reaction
C+02 = CO2
Required
The equilibrium constant
Solution
The equilibrium constant is the ratio of concentration or pressure between the product and the reactant with each reaction coefficient raised
The equilibrium constant is based on the concentration (Kc) in a reaction
pA + qB -----> mC + nD
![\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7B%5Cbold%20%7BKc%20~%20%3D%20~%20%5Cfrac%20%7B%5BC%5D%20%5E%20m%20%5BD%5D%20%5E%20n%7D%20%7B%5BA%5D%20%5E%20p%20%5BB%5D%20%5E%20q%7D%7D%7D%7D)
So for the reaction :
C+O₂ ⇔ CO₂
![\tt Kc=\dfrac{[CO_2]}{[C][O_2]}](https://tex.z-dn.net/?f=%5Ctt%20Kc%3D%5Cdfrac%7B%5BCO_2%5D%7D%7B%5BC%5D%5BO_2%5D%7D)
There are 4 moles of spectator ions that remain in solution.
The equation of the reaction is;
Na2CO3(aq) + Pb(NO3)2(aq) -------> PbCO3(s) + 2NaNO3(aq)
We have to determine the limiting reactant. This is the reactant that yields the least amount of product. Note that the spectator ions are Na^+ and NO3^- that form NaNO3.
For Na2CO3
1 mole of Na2CO3 yields 2 moles of NaNO3
3 moles of Na2CO3 yields 3 × 2/1 = 6 moles of NaNO3
For Pb(NO3)2
1 mole of Pb(NO3)2 yields 2 moles of NaNO3
2 moles of Pb(NO3)2 yields 2 × 2/1 = 4 moles of NaNO3
We can see that Pb(NO3)2 is the limiting reactant.
Since [NaNO3] = [Na^+] = [NO3^-], it follows that there are 4 moles of spectator ions that remain in solution.
Learn more: brainly.com/question/22885959
Answer:
carbon
Explanation:
Atomic radius represents the distance from the nucleus to the outer shell of an element.