Answer:
D) Adding a catalyst
Explanation:
Adding a catalyst decreases activation energy and allows the reaction to occur more easily.
<span>11.3 kPa
The ideal gas law is
PV = nRT
where
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant (8.3144598 L*kPa/(K*mol) )
T = Absolute temperature
We have everything except moles and volume. But we can calculate moles by starting with the atomic weight of argon and neon.
Atomic weight argon = 39.948
Atomic weight neon = 20.1797
Moles Ar = 1.00 g / 39.948 g/mol = 0.025032542 mol
Moles Ne = 0.500 g / 20.1797 g/mol = 0.024777375 mol
Total moles gas particles = 0.025032542 mol + 0.024777375 mol = 0.049809918 mol
Now take the ideal gas equation and solve for P, then substitute known values and solve.
PV = nRT
P = nRT/V
P = 0.049809918 mol * 8.3144598 L*kPa/(K*mol) * 275 K/5.00 L
P = 113.8892033 L*kPa / 5.00 L
P = 22.77784066 kPa
Now let's determine the percent of pressure provided by neon by calculating the percentage of neon atoms. Divide the number of moles of neon by the total number of moles.
0.024777375 mol / 0.049809918 mol = 0.497438592
Now multiply by the pressure
0.497438592 * 22.77784066 kPa = 11.33057699 kPa
Round the result to 3 significant figures, giving 11.3 kPa</span>
Answer:
2HI + K2SO3=>2KI+H2SO3
Explanation:When aqueous hydroiodic acid and aqueous potassium sulfite are mixed the products obtained are potassium iodide and sulfurous acid.Both reactants are ionic compounds and they undergo double replacement reaction.In a double replacement reaction the parts of the ionic compounds are changed.The product is obtained by combinig cation of one compound with anion of other compound.so in above reaction sulfurous acid is obtained which is in gaseous form and potassium iodide is an ionic compound.
D- viscosity has no effect on boiling point
