62.5 mL is prepare .
What is molarity?
Molar concentration is a unit of measurement for the concentration of a chemical species, specifically a solute, in a solution, expressed as the amount of substance per unit volume of solution. The most often used unit for molarity in chemistry is the number of moles per litre, denoted by the unit symbol mol/L or mol/dm3 in SI units.
Molarity of the stock solution as 0.100 M
Volume of the dilute solution as 250 mL
Molarity of dilute solution as 0.0250 M
We are required to calculate the Volume of the stalk solution.
Taking the volume and molarity of the stock solution to be V₁ and M₁ respectively, and volume and molarity of the dilute solution to be V₂ and M₂ respectively.
We are going to use the dilution formula;
According to the dilution formula, M₁V₁ = M₂V₂
Rearranging the formula;
V₁ = M₂V₂ ÷ M₁
= (0.025 M × 0.25 L) ÷ 0.100 M
= 0.0625 L
But, 1 L = 1000 mL
V₁ = 62.5 mL
Therefore, the volume of the stock solution is 62.5 mL
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Answer:
The molecules inside the ice pack are frozen therefore they are stuck in place. The molecules inside an insulated cup aren't frozen but still stuck inside of their container so that they can not fly out.
Explanation:
Answer:c
Explanation:
A valley will be carved out due to erosion.
Answer:
B - What we change
Explanation:
Dependent Variable - What we measure
Control Variable - what stays the same
Conclusion - what we conclude
<em>Hope</em><em> </em><em>this</em><em> </em><em>can</em><em> </em><em>Help</em><em>!</em>
<em>:</em><em>D</em>
Answer:
ΔH = +155.6 kJ
Explanation:
The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).
If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:
N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ
2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ
2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ
The intermediares are N₂ and O₂, thus, reorganizing the reactions:
N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ
NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)
N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)
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N₂O(g) + NO₂(g) → 3NO(g)
ΔH = +180.7 + 56.55 - 81.6
ΔH = +155.6 kJ
