At 365 K temperature sulfur tetrafluoride have a density of 0.260 g/L at 0.0721 atm.
<h3>What is an ideal gas equation?</h3>
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
First, calculate the moles of the gas using the gas law,
PV=nRT, where n is the moles and R is the gas constant. Then divide
the given mass by the number of moles to get molar mass.
Given data:
P= 0.0721 atm
T=?
Putting value in the given equation:
T = 365.2158727 K= 365 K
Hence , at 365 K temperature sulfur tetrafluoride have a density of 0.260 g/L at 0.0721 atm.
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<em>Acetic acid, HC2H3O2</em>
First, calculate for the molar mass of acetic acid as shown below.
M = 1 + 2(12) + 3(1) + 2(16) = 60 g
Then, calculating for the percentages of each element.
<em> Hydrogen:</em>
P1 = ((4)(1)/60)(100%) = <em>6.67%</em>
<em> Carbon:</em>
P2 = ((2)(12)/60)(100%) = <em>40%</em>
<em>Oxygen</em>
P3 =((2)(16) / 60)(100%) = <em>53.33%</em>
<em>Glucose, C6H12O6</em>
The molar mass of glucose is as calculated below,
6(12) + 12(1) + 6(16) = 180
The percentages of the elements are as follow,
<em> Hydrogen:</em>
P1 = (12/180)(100%) = <em>6.67%</em>
<em>Carbon:</em>
P2 = ((6)(12) / 180)(100%) = <em>40%</em>
<em>Oxygen:</em>
P3 = ((6)(16) / 180)(100%) = <em>53.33%</em>
b. Since the empirical formula of the given substances are just the same and can be written as CH2O then, the percentages of each element composing them will just be equal.
Answer:
See explanation
Explanation:
The reaction equation is;
C3H8 (g) + 5O2(g) -------> 4H2O(g) + 3CO2(g)
From the formula;
Total enthalpy of reactants = (ΔHf of Reactant 1 x Coefficient) + (ΔHf of Reactant 2 x Coefficient)
Total enthalpy of products= (ΔHf of Product 1 x Coefficient) + (ΔHf of Product 2 x Coefficient)
Hence;
Total enthalpy of reactants =[(-103.85 * 1) + (0 * 5)] = -103.85 + 0 = -103.85 KJ/mol
Total enthalpy of products= [(-393.51 * 4) +(-241.82 * 3)] = (-1574.04) + (-483.64) = -2057.68 KJ/mol