Cocaine selectively blocks sodium channels which is the mechanism that leads to local anesthetic effects.
Voltage gated sodium channels play very important roles in the body as they are responsible for action initiation and propagation in excitable cells, such as nerves, muscles and neuroendocrine cells. Like other sodium ion channels blockers such as lidocaine, Cocaine selectively blocks sodium ion channels which denies entry of sodium ions in the cell, thus leading to local anesthetic effects.
Answer:
D) Carbon Dioxide
Explanation:
Carbon is a greenhouse gas that traps heat.
Carbon is in Carbon Dioxide **Obviouslyyy lol
Answer: 8moles
Explanation:
The reaction below shows the formation of 2 moles of water from 2 moles of hydrogen and 1 mole of oxygen respectively.
2H2(g) + O2 (g) --> 2H2O(l)
So, if 1 mole of O2 produce 2 mole of H2O
4 moles of O2 will produce Z mole of H2O
To get the value of Z, cross multiply
1 x Z = 4 x 2
Z = 8
So, the equation will be 8H2(g) + 4O2 (g) --> 8H2O(l)
Thus, 4 moles of O2 will produce 8moles of H2O .
Answer:
sun, air, water, wind and ocean current
Explanation:
As sun is the major source of energy on earth, the earth gets heated due to solar energy, this creates convection currents in the Air and Water, the heating of these areas create wind and ocean currents due to difference in temperature and pressure.
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)
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