The noble gases are helium, neon, argon, krypton, xenon, radon, and ununoctium. The noble gases are relatively nonreactive. This is because they have a completevalence shell<span>. They have little tendency to </span>gain<span> or lose </span>electrons. <span>These gases all have similar properties under standard conditions: they are all odorless, colorless, monatomic gases with very low chemical </span>reactivity<span>. The six noble gases that occur naturally are helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), and Radon (Rn).</span>
Answer:
pH = 10.9
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to say that the undergoing reaction between this buffer and OH⁻ promotes the formation of more CO₃²⁻ because it acts as the base, we can do the following:
The resulting concentrations are:
![[CO_3^{2-}]=\frac{0.1435mol}{0.25L}=0.574M \\](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D%5Cfrac%7B0.1435mol%7D%7B0.25L%7D%3D0.574M%20%5C%5C)
![[HCO_3^{-}]=\frac{0.0265mol}{0.25L}=0.106M](https://tex.z-dn.net/?f=%5BHCO_3%5E%7B-%7D%5D%3D%5Cfrac%7B0.0265mol%7D%7B0.25L%7D%3D0.106M)
Thus, since the pKa of this buffer system is 10.2, the change in the pH would be:
Which makes sense since basic OH⁻ ions were added.
Regards!
The solution would be like
this for this specific problem:
<span><span>
E</span>=</span><span>mc</span>ΔT<span> <span>
= (</span>15<span> g</span><span>)(</span>1.91<span> <span>J<span><span>g∘</span>C</span></span>)(</span>25<span><span> ∘</span>C</span>−15<span><span> ∘</span>C</span><span>)
</span></span>= 28.65 * 10
= 286.5
<span>
I hope this helps and if you have any further questions, please don’t hesitate
to ask again. </span>
Answer:
= 24 square units
Explanation:
The area of a rhombus is given by the formula;
Area = pq/2
Where p and q are the diagonals of the rhombus.
In this case; p = 6 and q = 8
Therefore;
Area = (6×8)/2
= 24 square units
