I’m pretty sure it’s in a group (column) of the period table. Hope this helps :)))
Ca(OH)₂ ==> Ca²⁺ + 2 OH<span>-
Ca(OH)</span>₂ is <span>strong Bases</span><span>
</span>Therefore, the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻<span>.
</span>
pOH = - log[ OH⁻]
pOH = - log [ <span>5 x 10⁻⁴ ]
pOH = 3.30
pH + pOH = 14
pH + 3.30 = 14
pH = 14 - 3.30
pH = 10.7
hope this helps!</span>
Answer:
Idk if this is right but i hope it helps... sorry if it's wrong
Explanation:
Answer:
60 V
Explanation:
From;
Vs/Vp = Ns/Np
Where;
Vs = voltage in the secondary coil = 6V
Vp = voltage in the primary coil= ??
Ns = number of turns in the secondary coil = 9
Np= number of turns in the primary coil = 90
6/Vp = 9/90
Vp= 90 * 6/9
Vp= 60 V
<span>PV/T = P'V'/T'
660 x 1.00/295.2 = P' x 1.00/317.8
P'=710.5 torr</span>
