Answer:
Reacciones de neutralización
Mg(OH)2
NaHCO3
Bicarbonato de sodio
Cloruro de sodio
Reacciones acido base
HCl + NaOH → NaCl + H2O.
H2SO4 + 2 NaOH → Na2SO4 + H2O.
HCl + NH3 → NH4Cl + H2O.
HCN + NaOH → NaCN + H2O.
Explanation:
Answer:
No it does not i have tried it once before and it would not load for me and it froze the whole page site. But it might be different for you...
Explanation:
Answer:
[OH-] = 6.17 *10^-10
Explanation:
Step 1: Data given
pOH = 9.21
Step 2: Calculate [OH-]
pOH = -log [OH-] = 9.21
[OH-] = 10^-9.21
[OH-] = 6.17 *10^-10
Step 3: Check if it's correct
pOH + pH = 14
[H+]*[OH-] = 10^-14
pH = 14 - 9.21 = 4.79
[H+] = 10^-4.79
[H+] = 1.62 *10^-5
6.17 * 10^-10 * 1.62 * 10^-5 = 1* 10^-14
The answer is 1) CF3
Because: the equivalent of Fluorine is -1 so it aims to get an electron so much. which means it's electronegative.
but the equivalent of Hydrogen is +1 so it aims to give an electron. which means it's electropositive.
please mark as brainliest answer