Explanation:
The given data is as follows.
Concentration = 0.1 
= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions
= 
T =
= (30 + 273) K = 303 K
Formula for electric double layer thickness (
) is as follows.
= 
where,
= concentration = 
Hence, putting the given values into the above equation as follows.
=
=
=
m
or, =
= 1 nm (approx)
Also, it is known that
= 
Hence, we can conclude that addition of 0.1
of KCl in 0.1
of NaBr "
" will decrease but not significantly.
Answer:
Explanation:
a. CuO+ 2HCl⇒CuCl2+ H2O
b.
=
= 0,05 (mol)
⇒
=
=0,05 mol
⇒
= 0,05×135=6,75 (g)
c.
=2×
=0,1 (mol)
⇒
= 0,1×36,5= 3,65 (g)
⇒
=
×100=36,5 (g)
⇒ Nồng độ phần trăm dd sau phản ứng= Nồng độ % dd CuCl2=
×100=
×100≈ 16,67%
Answer:
mendeleev left a space
Explanation:
so the periodic table can be organize
It B I just took the test ;)
Answer:
To calculate the number we need molar mass.