The ideal gas law may be written as

where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)
For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol
Therefore

Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
= 221.95 kPa
= (2.295 x 10⁵)/101325 atm
= 2.19 atm
Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)
Answer:
The molarity is 2M
Explanation:
First , we calculate the weight of 1 mol of NaCl:
Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol
58,5 g---1 mol NaCl
233,772 g--------x= (233,772 g x1 mol NaCl)/58,5 g= 4 mol NaCl
<em>A solution molar--> moles of solute in 1 L of solution:</em>
2 L-----4 mol NaCl
1L----x0( 1L x4mol NaCl)/4L =2moles NaCl---> 2 M
It will received an H+. I’m not sure if that answers your question correctly
Mass number...use da textbook bra