<span><em>Answer:</em>
A strontium-90 atom that has a lost two electrons has <u>38</u> protons, <u>52</u> neutrons, and <u>36</u> electrons.
<em>Explanation:
</em>Atomic number<em> of </em>Strontium (Sr) is 38.
<em>Atomic number = number of protons
</em>Hence, Strontium has 38 protons.
If the element is in neutral state,
number of protons = number of electrons.
Then, neutral Strontium atom should have 38 electrons.
But the question says Sr has lost 2 electrons. Hence, number of electrons should be 38 - 2 = 36.
Mass number = number of protons + number of neutrons.
The given mass number is 90. Hence, number of neutrons should be 90 - 38 = 52.</span>
Answer:
Elements only contain one type of atoms while compounds contain two or more types of atoms.
Explanation:
An example of an element is sodium --> Na (only Na atoms)
An example of a compound can be water --> H2O (contain H and O atoms)
*But the particles within a compound are all the same.
The answer is (2) 2 pairs. The oxygen atoms combine to get stable structure. So after the combination, each atom needs to have 8 electrons. So when they share 4 electrons can satisfy this requirement. Then there are 2 pairs shared.
Answer:
pH=11.
Explanation:
Hello!
In this case, since the data is not given, it is possible to use a similar problem like:
"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"
Thus, for the reaction:

Tt is possible to compute the remaining moles of ethylamine via the following subtraction:

Thus, the concentration of ethylamine in solution is:
![[ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M](https://tex.z-dn.net/?f=%5Bethylamine%5D%3D%5Cfrac%7B0.0816mol%7D%7B0.1850L%2B0.1144L%7D%3D0.2725M)
Now, we can also infer that some salt is formed, and has the following concentration:
![[salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M](https://tex.z-dn.net/?f=%5Bsalt%5D%3D%5Cfrac%7B0.0549mol%7D%7B0.1850L%2B0.1144L%7D%3D0.1834M)
Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:
![pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0](https://tex.z-dn.net/?f=pOH%3DpKb%2Blog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D%20%29%5C%5C%5C%5CpOH%3D3.19%2Blog%28%5Cfrac%7B0.1834M%7D%7B0.2725M%7D%29%5C%5C%5C%5CpOH%3D3.0)
Finally, the pH turns out to be:

NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.
Best regards!
A. More reactants could have been added to reaction B