The partial pressure of Hydrogen gas can directly be calculated
by simply taking the difference of the overall pressure and the vapour pressure
of water. That is:
P (H2 gas) = 759.2 torr – 23.8 torr
<span>P (H2 gas) = 735.4 torr</span>
Solute - the substance that dissolves into the solvent to produce a homogenous mixture
Solvent - the substance in which a solute dissolves to produce a homogeneous mixture
Answer:
Explanation:
alright, dawg, lets get this bread. CHEMISTRY? OH YEAH I LOVE CHEMISTRY.
what is a mol? do you know who avogadro is? sounds like avocado. free shavocado. ok so you MUST REMEMBER THIS NUMBER PLEASE.
please remember this number and commit it to your memory: avogadros number

this is how much a mole is. you know how a pair is 2 and a dozen is 12? ok so a mole is
it is confusing at first but hopefully this helps you to understand.
now that we understand this..... lets perform this calculation with a calculator

notice i divide the question by the avogadros number to find out how many moles are in the number. ok but listen... it gets into a tough area here... because HOW ARE WE TO DIVIDE SUCH A HUMONGOUS NUMBER BY ANOTHER HUMONGOUS NUMBER?!?!?
its easy, its cake, just listen this is how you do it. only focus on the numbers NOT the 10 exponential ones. so just 3.90 and 6.02 ok? lets divide these two numbers 3.90 / 6.02 and we get 0.6478... how interesting... ok now lets deal with the exponents of 10. notice that we are DIVIDING these numbers so think of it as subtracting the exponents of ten..... 22 minus 23 equals -1
so we have 
now this negative 1 thing is annoying so lets just make it to the power of 0

and anything to the power of 0 just becomes 1.
0.06478
so this is our answer but keep in mind we need 3 sig figs. if we round then we get 0.0648
put this into scientific notation we get 
Combined gas law is
PV/T = K (constant)
P = Pressure
V = Volume
T = Temperature in Kelvin
For two situations, the combined gas law can be applied as,
P₁V₁ / T₁ = P₂V₂ / T₂
P₁ = 3.00 atm P₂ = standard pressure = 1 atm
V₁ = 720.0 mL T₂ = standard temperature = 273 K
T₁ = (273 + 20) K = 293 K
By substituting,
3.00 atm x 720.0 mL / 293 K = 1 atm x V₂ / 273 K
V₂ = 2012.6 mL
hence the volume of gas at stp is 2012.6 mL
Answer: grams=0.048g, ounces=0.0017oz, 0.00011lb
Explanation:
Stoichiometry
48 mg x 1 g
÷ 1000 mg = 0.048 g
48 mg x 1 g x 16 oz
÷ 1000 mg ÷ 453.6 g = 0.0017 oz
48 mg x 1 g x 1 lb
÷ 1000 mg ÷ 453.6 g = 0.00011 lb