Explanation:
The number of moles of solute present in liter of solution is defined as molarity.
Mathematically, Molarity = 
Also, when number of moles are equal in a solution then the formula will be as follows.

It is given that
is 8.00 M,
is 7.00 mL, and
is 0.80 M.
Hence, calculate the value of
using above formula as follows.



= 70 ml
Thus, we can conclude that the volume after dilution is 70 ml.
Answer:
- <em>The volume of 14.0 g of nitrogen gas at STP is </em><u><em>11.2 liter.</em></u>
Explanation:
STP stands for standard pressure and temperature.
The International Institute of of Pure and Applied Chemistry, IUPAC changed the definition of standard temperature and pressure (STP) in 1982:
- Before the change, STP was defined as a temperature of 273.15 K and an absolute pressure of exactly 1 atm (101.325 kPa).
- After the change, STP is defined as a temperature of 273.15 K and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).
Using the ideal gas equation of state, PV = nRT you can calculate the volume of one mole (n = 1) of gas. With the former definition, the volume of a mol of gas at STP, rounded to 3 significant figures, was 22.4 liter. This is classical well known result.
With the later definition, the volume of a mol of gas at STP is 22.7 liter.
I will use the traditional measure of 22.4 liter per mole of gas.
<u>1) Convert 14.0 g of nitrogen gas to number of moles:</u>
- n = mass in grams / molar mass
- Atomic mass of nitrogen: 14.0 g/mol
- Nitrogen gas is a diatomic molecule, so the molar mass of nitrogen gas = molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol
- n = 14.0 g / 28.0 g/mol = 0.500 mol
<u>2) Set a proportion to calculate the volume of nitrogen gas:</u>
- 22.4 liter / mol = x / 0.500 mol
- Solve for x: x = 0.500 mol × 22.4 liter / mol = 11.2 liter.
<u>Conclusion:</u> the volume of 14.0 g of nitrogen gas at STP is 11.2 liter.
Answer:
A
Explanation:
BF3 has a trigonal planar shape.
XeO3 has a trigonal pyramidal shape.
The question is incomplete, complete question is ;
A deep-sea diver uses a gas cylinder with a volume of 10.0 L and a content of 51.8 g of
and 33.1 g of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21°C.Express the pressures in atmospheres to three significant digits separated by commas.
Answer:
Partial pressure of the oxygen gas is 3.91 atm.
Partial pressure of the helium gas is 20.0 atm
Total pressure of the gases is 24.0 atm
Explanation:
Moles of oxygen gas = 
Moles of helium gas = 
Total moles of gas = 
Volume of the cylinder = V = 10.0 L
Total pressure in the cylinder = P = ?
Temperature of the gas in cylinder = T = 21°C = 21 + 273 K = 294 K
PV = nRT ( ideal gas equation )


P = 23.88 atm ≈ 23.9
Partial pressure of the individual gas will be determined by the help of Dalton's law:
partial pressure = Total pressure × mole fraction of gas
Partial pressure of the oxygen gas


Partial pressure of the helium gas

