H^+(aq) + OH^-(aq) ---> H2O(l)
<span>Na^+ and ClO4^- are the spectator ions.</span>
<span>the bonds in iron(III) oxide are more ionic</span>
Answer:
0.7457 g is the mass of the helium gas.
Explanation:
Given:
Pressure = 3.04 atm
Temperature = 25.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (25.0 + 273.15) K = 298.15 K
Volume = 1.50 L
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
3.04 atm × 1.50 L = n × 0.0821 L.atm/K.mol × 298.15 K
<u>⇒n = 0.1863 moles</u>
Molar mass of helium = 4.0026 g/mol
The formula for the calculation of moles is shown below:
Thus,
<u>0.7457 g is the mass of the helium gas. </u>
<span>2.10 grams.
The balanced equation for the reaction is
CO + 2H2 ==> CH3OH
The key thing to take from this equation is that it takes 2 hydrogen molecules per carbon monoxide molecule for this reaction. And since we've been given an equal number of molecules for each reactant, the limiting reactant will be hydrogen.
We can effectively claim that we have 5.86/2 = 2.93 l of hydrogen and an excess of CO to consume all of the hydrogen. So the number of moles of hydrogen gas we have is:
2.93 l / 22.4 l/mol = 0.130803571 mol
And since it takes 2 moles of hydrogen gas to make 1 mole of methanol, divide by 2, getting.
0.130803571 mol / 2 = 0.065401786 mol
Now we just need to multiply the number of moles of methanol by its molar mass. First lookup the atomic weights involved.
Atomic weight carbon = 12.0107 g/mol
Atomic weight hydrogen = 1.00794 g/mol
Atomic weight oxygen = 15.999 g/mol
Molar mass CH3OH = 12.0107 + 4 * 1.00794 + 15.999 = 32.04146 g/mol
So the mass produced is
32.04146 g/mol * 0.065401786 mol = 2.095568701 g
And of course, properly round the answer to 3 significant digits, giving 2.10 grams.</span>
N=3^K-1 I really don't know how to explain it but that's the formula <span />
