To answer this, we must first know what STP is. STP (Standard Temperature and Pressure) is when the environment is at the following:
To figure out any missing units in a gas equation, we use the Ideal Gas Law, PV=nRT.
To work out the number of mol, we substitute in the calues we have and solve for n. This gives us 101.325kPa*5.6dm3=n*8.314*273K. Therefore, n=0.25 mol.
Alternatively, we could simply use the fact that 1 mol of any gas at STP takes up 22.4 litres. So 5.6/22.4 leaves us with 0.25 mol.
There are 0.25 moles of argon gas present in 5.6 liters at standard conditions (1 mole of any gas occupies 22.4 liters at standard conditions).
Molarity = moles of solute/Liters of solution (40 ml = 0.04 Liters) algebraically manipulated, Moles of solute = Liters of solution * Molarity Moles HCl = (0.04 Liters)(0.035 M) = 0.0014 moles HCl ==============
Amount of HCl = 700/1000 x 0.33 = 0.231 moles
Molarity = moles of solute/Liters of solution ( 75.0 ml = 0.075 Liters ) Algebraically manipulate, moles of solute = Liters of solution * Molarity Moles KMnO4 = (0.075 Liters)(0.0950 M) = 7.13 X 10 -3 moles KMnO4 ------------------------------------
To find the number of moles of argon in 37.9 g, you need to use the molar mass of argon, which is 39.95 g/mol. Divide the given mass by the molar mass to get the number of moles. So, 37.9 g / 39.95 g/mol = 0.95 moles of argon.
There are 0.25 moles of argon gas present in 5.6 liters at standard conditions (1 mole of any gas occupies 22.4 liters at standard conditions).
8,4 liters of nitrous oxide at STP contain 2,65 moles.
The number of moles of helium is 0,32.
The answer is 13,89 moles.
Molarity = moles of solute/Liters of solution (40 ml = 0.04 Liters) algebraically manipulated, Moles of solute = Liters of solution * Molarity Moles HCl = (0.04 Liters)(0.035 M) = 0.0014 moles HCl ==============
Amount of HCl = 700/1000 x 0.33 = 0.231 moles
To determine the number of moles of KCl present in a solution, you need to use the formula: moles = molarity x volume (in liters). First, convert the volume from milliliters to liters by dividing by 1000 (50.0 mL = 0.050 L). Then, calculate the moles of KCl by multiplying the molarity (0.552 M) by the volume in liters (0.05 L). This gives you approximately 0.0286 moles of KCl in the solution.
1,67.1024 argon atoms is equal to 2,773 moles.
Molarity = moles of solute/Liters of solution ( 75.0 ml = 0.075 Liters ) Algebraically manipulate, moles of solute = Liters of solution * Molarity Moles KMnO4 = (0.075 Liters)(0.0950 M) = 7.13 X 10 -3 moles KMnO4 ------------------------------------
To find the number of moles of argon in 37.9 g, you need to use the molar mass of argon, which is 39.95 g/mol. Divide the given mass by the molar mass to get the number of moles. So, 37.9 g / 39.95 g/mol = 0.95 moles of argon.
The answer is approx. 2 moles (for anhydrous sodium sulfate).
At STP, 1 mole of any gas occupies 22.4 L. Therefore, in a 5L sample of argon at STP, there would be 5/22.4 moles of argon, which is approximately 0.223 moles.