Answer:
10 atm
Explanation:
There's a lot to do here, but lets take it one step at a time. First, let's write a balanced equation for the decomposition of potassium chlorate into potassium chloride and oxgyen gas.
2 KClO3 → 2 KCl + 3 O2
Now let's find the moles of the KClO3 (molar mass 122.55 g/mol) that we have take 10 g/122.55 g/mol, grams will cancel and we are left with 0.0816 moles. lets divide that by two since we have a two in front of the KClO3 in the equation, and then multiply that number by 5 since it's the total moles of products, in summary, multiply by 5/2 to get 0.204 moles.
Now that we know the moles of our products, let's plug some stuff into the ideal gas law PV = nRT. We are looking for P so let's solve for that. P = (nRT)/V, now let's plug in our values. Make sure V is converted to liters so 0.5 L. And convert celcius to kelvin by adding 273
P = ((0.204 moles)(318 K)(0.08206 L atm mol^-1 K^-1))/0.5 L
A lot of units cancel, and we get about 10.65 atm, if you don't want the answer in atm, you can find a conversion equation. But let's round to sig figs for now, which will bring us to 10 atm.
Answer:
A conjugate acid, is a species formed by the reception of a proton (H+) by a base—in other words, it is a base with a hydrogen ion added to it.
OH- is the conjugate base of H2O..
Answer: The 3rd and 6th bullet point is the quantitative data.
Explanation: Quantitative data is expressed by NUMBERS and Qualitative data is expressed by WORDS. The 3rd and 6th one is correct because they both use numbers to compare how much time hummingbirds spent feeding on nectar.
The combustion reaction of propane would be expressed as:
C3H8 + 5O2 = 3CO2 + 4H2O
To determine the mass of water that is produced from the given amount of propane, we use the mass of propane and the relation of the substances from the balanced reaction. We do as follows:
moles propane = 22 g C3H8 ( 1 mol / 44.1 g ) = 0.50 mol C3H8
moles H2O = 0.50 mol C3H8 ( 4 mol H2O / 1 mol C3H8) = 2 mol H2O
mass H2O = 2 mol H2O ( 18.02 g / 1 mol ) = 36.04 g H2O
Therefore, the mass of water that is produced from 22 grams of propane would be 36.04 g.
