Answer:
The total pressure is 27.8 atm
Explanation:
From the ideal gas equation,
PV = nRT
P (total pressure) = nRT/V
n (total moles of gases) = (6/1 moles of hydrogen) + (15.2/14 moles of nitrogen) + (16.8/4 moles of helium) = 6+1.1+4.2 = 11.3 moles
R = 0.082057L.atm/gmol.K, T = 27°C = 27+273K = 300K, V = 10L
P = 11.3×0.082057×300/10 = 27.8 atm
Hello!
Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).
We have the following data:
m (mass) = ?
n (number of moles) = 5.20 moles
MM (Molar mass of C6H12) ≈ 84.2 g/mol
Now, let's find the mass, knowing that:
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I Hope this helps, greetings ... Dexteright02! =)
Answer:
6.4 g BaSO₄
Explanation:
You have been given the molarity and the volume of the solution. To find the mass of the solution, you need to (1) find the moles BaSO₄ (via the molarity ratio) and then (2) convert moles BaSO₄ to grams BaSO₄ (via the molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given values.
Molarity (mol/L) = moles / volume (L)
(Step 1)
55 mL / 1,000 = 0.055 L
Molarity = moles / volume <----- Molarity ratio
0.5 (mol/L) = moles / 0.055 L <----- Insert values
0.0275 = moles <----- Multiply both sides by 0.055
(Step 2)
Molar Mass (BaSO₄): 137.33 g/mol + 32.065 g/mol + 4(15.998 g/mol)
Molar Mass (BaSO₄): 233.387 g/mol
0.0275 moles BaSO₄ 233.387 g
--------------------------------- x ------------------- = 6.4 g BaSO₄
1 mole
If you know the Table of elements you can see it on.
