There are 90.0 g of F are present in 185 g of CaF₂. In chemistry, a mole, usually spelled mol, is a common scientific measurement unit for significant amounts of extremely small objects like atoms, molecules, or other predetermined particles.
According to question, mass of CaF₂ = 185 g
It is required to calculate the moles of CaF₂
Moles of CaF₂ = 185 g / 78.074 g.mol-1
= 2.369 mole of CaF₂
Now find the moles of F from the moles of CaF₂
1 mole of CaF₂ = 2 moles of F
2.369 mole of CaF₂ = ?
= 4.74 moles of F
Now change the mole to gram of F
Mass of F = 4.74 moles of F × 18.998 g/mol
= 90.03 g of F
Thus, 90.0 g of F are in 185 g of CaF₂.
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The thanksgiving luang cave system is estimated to be 35 million cubic meters large. An estimated 250 million liters of water was pumped out of the Dave over the course of the rescue if all the water had remained in the cave system instead of being pumped out what percent of the cave would have been filled with water?
If all the pumped water had remained in the cave system, approximately 71.43% of the cave would have been filled with water.
To calculate the percentage of the Thanksgiving Luang cave system that would have been filled with water if all the pumped water had remained, we need to find the ratio of the volume of water to the total volume of the cave system.Given that the cave system is estimated to be 35 million cubic meters large and 250 million liters of water were pumped out, we need to convert liters to cubic meters. Since 1 liter is equal to 0.001 cubic meters, the pumped water volume in cubic meters is 250 million multiplied by 0.001, which is 250,000 cubic meters.To find the percentage, we divide the volume of water (250,000 cubic meters) by the total volume of the cave system (35 million cubic meters) and multiply by 100.
Percentage = (Volume of water / Total volume of cave system) x 100
= (250,000 / 35,000,000) x 100
≈ 0.7143 x 100
≈ 71.43%
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For the reaction
4PH3(g)↽−−⇀6H2(g)+P4(g)
the equilibrium concentrations were found to be [PH3]=0.250 M, [H2]=0.580 M, and [P4]=0.750 M.
What is the equilibrium constant for this reaction?
c=
The equilibrium constant for the given reaction is approximately 25.199.
To determine the equilibrium constant for the given reaction, we can use the concentrations of the species at equilibrium. The equilibrium constant (Kc) is defined as the ratio of the product concentrations raised to their stoichiometric coefficients divided by the ratio of the reactant concentrations raised to their stoichiometric coefficients.
The balanced equation for the reaction is:
4PH3(g) ↔ 6H2(g) + P4(g)
Using the concentrations given at equilibrium, we can substitute the values into the equilibrium expression:
Kc = ([H2]^6 * [P4]) / ([PH3]^4)
Plugging in the given concentrations:
Kc = ([0.580 M]^6 * [0.750 M]) / ([0.250 M]^4)
Kc = (0.1311 * 0.750) / (0.00390625)
Kc ≈ 25.199
This value indicates that at equilibrium, the product concentrations are favored over the reactant concentrations, as Kc is greater than 1.
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What volume of CO2(g), measured at STP is produced if 15.2 grams of CaCO(s) is heated?
Answer:
Volume = 3.4 L
Explanation:
In order to calculate the volume of CO₂ produced when 15.2 g of CaCO₃ is heated, we need to first write out the balanced equation of the thermal decomposition of CaCO₃:
CaCO₃ (s) + [Heat] ⇒ CaO (s) + CO₂ (g)
Now, let's calculate the number of moles in 15.2 g CaCO₃:
mole no. = [tex]\mathrm{\frac{mass}{molar \ mass}}[/tex]
= [tex]\frac{15.2}{40.1 + 12 + (16 \times 3)}[/tex]
= 0.1518 moles
From the balanced equation above, we can see that the stoichiometric molar ratios of CaCO₃ and CO₂ are equal. Therefore, the number of moles of CO₂ produced is also 0.1518 moles.
Hence, from the formula for the number of moles of a gas, we can calculate the volume of CO₂:
mole no. = [tex]\mathrm{\frac{Volume \ in \ L}{22.4}}[/tex]
⇒ [tex]0.1518 = \mathrm{\frac{Volume}{22.4}}[/tex]
⇒ Volume = 0.1518 × 22.4
= 3.4 L
Therefore, if 15.2 g of CaCO₃ is heated, 3.4 L of CO₂ is produced at STP.
