Answer:
5.37 × 10⁻⁴ mol/L
Explanation:
<em>A chemist makes 660. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a 0.00154 mol/L stock solution of magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.</em>
Step 1: Given data
- Initial concentration (C₁): 0.00154 mol/L
- Initial volume (V₁): 230. mL
- Final concentration (C₂): ?
- Final volume (V₂): 660. mL
Step 2: Calculate the concentration of the final solution
We want to prepare a dilute solution from a concentrated one. We can calculate the concentration of the final solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
C₂ = C₁ × V₁ / V₂
C₂ = 0.00154 mol/L × 230. mL / 660. mL = 5.37 × 10⁻⁴ mol/L
Answer:
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Explanation:
The number of moles of NH3 that could be made would be 0.5 moles
<h3>Stoichiometric reactions</h3>
From the balanced equation of the reaction:
N2 (g) + 3 H2(g) ----> 2NH3 (g)
The mole ratio of N2 to H2 is 1:3
Thus, for 0.50 moles of N2, 1.5 moles of H2 should be present. But 0.75 moles of H2 was allowed to react. Meaning that H2 is limiting in this case.
Mole ratio of H2 and NH3 = 3:2
Thus for 0.75 moles H2, the mole of NH3 that would be produced will be:
2 x 0.75/3 = 0.5 moles
More on stoichiometric calculations can be found here: brainly.com/question/8062886
Answer:
6 moles of Cl2
Explanation:
First, the equation has to be balanced, which makes it 4 FeCl3 + 3 O2 --> 2 Fe2O3 + 6 Cl2
Using this information, we can see that one mole of O2 will not be present in the reaction. Since four moles of FeCl3 are needed to react in the equation, which would produce six moles of Cl2, and only four moles of FeCl3 are present, six moles of Cl2 would be produced.
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