Answer:
- 130.64°C.
Explanation:
- We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
<em>V₁T₂ = V₂T₁</em>
<em></em>
V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.
V₂ = 307.0 L, T₂ = ??? K.
<em>∴ T₂ = V₂T₁/V₁ </em>= (307.0 L)(294.0 K)/(634.0 L) = <em>142.36 K.</em>
<em>∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.</em>
Answer:
14,938 tablets.
Explanation:
WE divide:
1.21 kg / 81 mg
1.21 kg = 1,210 g = 1.210,000 mg.
So it is 1,210,000 / 81
= 14,938 tablets.
Answer:
There is 52.33 grams of water produced.
Explanation:
Step 1: Data given
Mass of propane burned = 32.00 grams
Molar mass of propane = 44.1 g/mol
Oxygen is in excess
Molar mass of water = 18.02 g/mol
Step 2: The balanced equation
C3H8 + 5O2 → 4H2O + 3CO2
Step 3: Calculate moles of propane
Moles of propane = mass propane / molar mass of propane
Moles of propane = 32.00 grams / 44.1 g/mol
Moles of propane = 0.726 moles
Step 4: Calculate moles of H2O
Propane is the limiting reactant.
For 1 mol of propane consumed, we need 5 moles of O2 to produce 4 moles of H2O and 3 moles of CO2
For 0.726 moles of propane we'll have 4*0.726 = 2.904 moles of H2O
Step 5: Calculate mass of H2O
Mass of H2O = moles of H2O * molar mass of H2O
Mass of H2O = 2.904 moles * 18.02 g/mol
Mass of H2O = 52.33 grams
There is 52.33 grams of water produced.
Answer:
This formula contains 12 oxygen atoms.
Answer:
https://www.edencsd.org/cms/lib/NY19000545/Centricity/Domain/38/Organic%20Answers.pdf
Explanation: