False..........
it has so many consequences
Answer:
d) a pure substance that can be separated into different elements by chemical means.
Explanation:
Hello,
Sodium chloride (NaCl) is a chemical compound composed by sodium and chlorine. They are chemically bonded through a ionic bond which means that the sodium transfers just one electron (valence electron) to the chlorine to form the bond. Now, if we want to separate them, we must induce a chemical process because bond breaking implies a chemical change, not physical. Thus, a typical way to separate them (break the bond) is via electrolysis in which an electric driving force induce the aforementioned breakdown.
Best regards.
Answer:
When Blue litmus is put in sulphuric acid it changes red .
And red litmus doesn't changes..
Conclusion Acid changes blue ltmus to red.
Answer:
pH = 2.21
Explanation:
Hello there!
In this case, according to the reaction between NaF and HCl as the latter is added to the buffer:
It is possible for us to see how more HF is formed as HCl is added and therefore, the capacity of this HF/NaF-buffer is diminished as it turns acid. Therefore, it turns out feasible for us to calculate the consumed moles of NaF and the produced moles of HF due to the change in moles induced by HCl:
Next, we calculate the resulting concentrations to further apply the Henderson-Hasselbach equation:
![[HF]=\frac{0.450mol}{1.0L} =0.450M](https://tex.z-dn.net/?f=%5BHF%5D%3D%5Cfrac%7B0.450mol%7D%7B1.0L%7D%20%3D0.450M)
![[NaF]=\frac{0.050mol}{1.0L} =0.050M](https://tex.z-dn.net/?f=%5BNaF%5D%3D%5Cfrac%7B0.050mol%7D%7B1.0L%7D%20%3D0.050M)
Now, calculated the pKa of HF:
We can proceed to the HH equation:
![pH=pKa+log(\frac{[NaF]}{[HF]} )\\\\pH=3.17+log(\frac{0.05M}{0.45M} )\\\\pH=2.21](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5BNaF%5D%7D%7B%5BHF%5D%7D%20%29%5C%5C%5C%5CpH%3D3.17%2Blog%28%5Cfrac%7B0.05M%7D%7B0.45M%7D%20%29%5C%5C%5C%5CpH%3D2.21)
Best regards!
Since it goes to completion, we can use:
Use NaMaVa =NbMbVb
where
Na= moles of H+ in H2SO4
Ma= molarity of the acid
Va= volume of the acid
Nb= moles of OH- in NaOH
Mb= molarity of the base
Vb= molarity of the base
plug in the numbers
(2 moles of H+) (Volume acid) (10.00mL acid) = (1 mole of OH-) (.248M base) (18.71mL base)
20 V = 4.64
V= .232 M
