<h2>The isotopes of an element all have the same __(atomic, mass) __number, but they have different __(atomic,mass)__numbers.</h2>
Explanation:
The isotopes of an element all have the same __atomic number __, but they have different __mass __numbers.
The isotopes have same atomic number that is :
- Same number of electrons
- Same number of protons
- same electronic configuration
- same valence electrons
- same valency
- same symbol
The isotopes have different mass number that is :
They differ in number of neutrons .
For example : Isotopes of hydrogen are : H₁¹ , H₁² , H₁³
isotopes of Oxygen is : O¹⁶ , O¹⁷, O¹⁸
Malleablity is a physical property so hence
D. Physical property is the answer
C5H12(g) + 8 O2(g) = 5 CO2(g) + 6H2O(g)
molar ratio of H2O to O = 6:8
<span>we have 0.200 moles of H2O so we need 0.200 * 8/6 moles of O2 = 0.267 moles of O2
Therefore, there will be 0.267 moles of O2 gas required to yield 0.2 moles of H20.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
<span>The answer to this question would be: A. all acid names end with "-ic"
Not all acid have "-id" suffix for their name. The example of this was already mentioned in the question as the sulfite ion will be called sulfurous acid. The "-ous" name is given to acid with lower oxidation state. Sulfuric acid will be the H2SO4 which have higher oxidation state.</span>
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!
