Answer:
0.0900 mol/L
Explanation:
<em>A chemist makes 330. mL of nickel(II) chloride working solution by adding distilled water to 220. mL of a 0.135 mol/L stock solution of nickel(II) chloride in water. Calculate the concentration of the chemist's working solution. Round your answer to significant digits.</em>
Step 1: Given data
- Initial concentration (C₁): 0.135 mol/L
- Initial volume (V₁): 220. mL
- Final concentration (C₂): ?
- Final volume (V₂): 330. mL
Step 2: Calculate the concentration of the final solution
We prepare a dilute solution from a concentrated one. We can calculate the concentration of the working solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
C₂ = C₁ × V₁/V₂
C₂ = 0.135 mol/L × 220. mL/330. mL = 0.0900 mol/L
Answer:
Therefore, none of the zeros is a significant figure and hence there is only one significant figure in this measurement in support G. The measurement is 0.7 minutes. Here again, zero is not a significant figure and therefore the only significant figure is one which is seven in support Edge.
Explanation:
Answer:
Fluorine azide or triazadienyl fluoride (FN3) is a yellow green gas composed of nitrogen and fluorine with formula FN3. It is counted as an interhalogen compound, as the azide functional group is termed a pseudohalogen. It resembles ClN3, BrN3, and IN3 in this respect.
Explanation:
Given reaction:
N2(g) + 3H2(g) → 2NH3(g)
The standard free energy change is given as:
ΔG° = ∑ nΔGf(products) - ∑ nΔGf(reactants)
= [2ΔGf(NH3(g))] - [ΔGf(N2(g)) + 3ΔGf(H2(g))]
= [2(-16.48)] - [ 1(0) + 3(0)] = -32.96 kJ
Ans: Free energy of the reaction is -32.96 kJ, i.e. reaction is spontaneous.
Answer:
9.82% of iron (II) will be sequestered by cyanide
Explanation:
We should first consider that Iron (II) and cyanide react to form the following structure:
[Fe(CN)₆]⁻⁴
Having considered this:
5.60 Lt Fe(II) 3.00x10⁻⁵ M ,this is, we have 5.60x3x10⁻⁵ = 1.68x10⁻⁴ moles of Fe⁺² (in 5.60 Lt)
Then , we have 9 ml NaCN 11.0 mM:
9 ml = 0.009 Lt
11.0 mM (milimolar) = 0.011 M (mol/lt)
So: 0.009x0.011 = 9.9x10⁻⁵ moles of CN⁻ ingested
As we now that the complex structure is formed by 1 Fe⁺² : 6 CN⁻ :
9.9x10⁻⁵ moles of CN⁻ will use 1.65x10⁻⁵ moles of Fe⁺² (this is, this amount of iron (II) will be sequestered
[(1.65x10⁻⁵ sequestred Fe⁺²)/(1.68x10⁻⁴ total available Fe⁺²)x100
% sequestered iron (II) = 9.82%
