Answer:
0.238 M
Explanation:
A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.
C₁.V₁ = C₂.V₂
C₁ × 24.00 mL = 0.220 M × 52.00 mL
C₁ = 0.477 M
The concentration of Pb(ClO₃)₂ is:
Answer:
Chemical compounds all have different melting points.
Explanation:
chemical compounds all have different freezing and boiling points. Different chemical compounds means they will have different chemical structures.
First, we write the balanced equation for this reaction:
2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂
From this equation, we see that there are 2 moles of potassium iodide required for each mole of lead (II) nitrate. Moreover, we may use the formula:
Moles = volume (in L) * molarity
We find the molar relation ship for KI : Pb(NO₃)₂ to be 2 : 1. So:
M₁V₁ = 2M₂V₂
V₁ = 2M₂V₂/M₁
V₁ = 2 * 0.112 * 0.155 / 0.2
V₁ = 0.1736 L
The volume required is 173.6 mL
Answer:
1000L
Explanation:
the 1 is a sig fig and since the 0 is between the 1 and 4 its also a significant number. to round them off you look at the 0,then look back at the 4 since its less than 5 u round down. then u replace the 43 with 0's
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