They all have the same number of electrons in their outer shell.
<span>Answer is: 2940 mL of
the HCL solution.</span>
c₁(HCl) = 10.0 M.
V₂(AgNO₃<span>) = ?.
c</span>₂(AgNO₃<span>) = 0.85 M.
V</span>₁(AgNO₃<span>) = 250 mL </span>÷ 1000 mL/L = 0.25 L.
<span>
c</span>₁<span> - original concentration of the solution, before it
gets diluted.
c</span>₂<span> - final concentration of the solution, after dilution.
V</span>₁<span> - volume to be diluted.
V</span>₂<span> - final volume after dilution.
c</span>₁ · V₁ = c₂ · V₂<span>.
V</span>₂(HCl) = c₁ · V₁ ÷ c₂.
<span>
V</span>₂(HCl) = 10 M · 0.25 L ÷ 0.85 M.
<span>
V</span>₂(HCl) = 2.94 L ·
1000 mL = 2940 mL.
The amount of HCl required for one experiment - 13.5 µl
the volume in terms of L - 13.5 x 10⁻⁶ L
the volume of HCl available - 0.250 L
since one experiment uses up - 13.5 x 10⁻⁶ L
then number of experiments - 0.250 L / 13.5 x 10⁻⁶ L = 1.8 x 10⁴ times
the experiment can be carried out 18000 times
Answer:
CasH52(1) + 38 O2(g) → 25 CO2(g) + 26 H2O(g)
Explanation:
There are 6.022*10^23 molecules in 1 mole of carbon
So how many will moles will be 7.87*20^7?
Let the required number of moles be ‘x’.
1 mole ———6.022*10^23
x moles———7.87*10^7
(Cross multiplication)
x=7.87*10^7/6.022*10^23
Therefore x=1.3*10^-16
