Dark is just absence of light like if the re container a and container b u nee to pour hcl from a to so a has hcl while b has absence of hcl
tell me if u need more examples.hope it helps
The hydronium and hydroxide concentrations of a solution that is 5.0 x 10-3 M H2SO4 is 2.7.
pH= -log[H+] - (i)
10^-3=H2So4
H+= 2×10-3
here ,
h2so4 ——— 2[H+] + so4^2-
thus [H+]= 2*10^(-3) because hydrogen ion has two moles
pH= -log[H+]
pH= -log(2×10^-3)
pH= 3-log2
pH= 3-log2pH= 2.7
The pH is 2.7
<h3>What is pH?</h3>
PH is the degree of alkalinity and acidicity in a solution.
Therefore, The hydronium and hydroxide concentrations of a solution that is 5.0 x 10-3 M H2SO4 is 2.7
Learn more about pH from the link below.
https://brainly.in/question/9937410
The reaction formula CH4 + 2O2 → CO2 + 2H2O shows the oxidation of 1 mole of CH4 (Methane) will yield 1 mole of CO2 (Carbon Dioxide). Since 1 mole of CH4 will weigh 12g (for the Carbon) + 4g (1g for each Hydrogen) = 16g, then 32g of CH4 will correspond to 32g / 16g/mole = 2 moles. Therefore the oxidation of 2 moles of CH4 will yield 2 moles of CO2.
<span>The student should
follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
<span>
1. First he should
calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>
</span>Molarity of the solution = 2.0 M<span>
Volume of solution which should be prepared = 1 L
Molarity =
number of moles / volume of the solution
Hence, number of moles in 1 L = 2 mol
2. Find
out the mass of dry CaCl</span>₂ in 2 moles.<span>
moles =
mass / molar mass
Moles of CaCl₂ =
2 mol</span><span>
Molar mass of CaCl₂ = </span><span>110.98 g/mol
Hence, mass of CaCl</span>₂ = 2 mol x <span>110.98 g/mol
= 221.96
g
3. Weigh the mass
accurately
4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and
finally wash the funnel and watch glass
with de-ionized water. That water also should be added into the volumetric
flask.
5. Then add some
de-ionized water into
the volumetric flask and swirl well until all salt are
dissolved.
<span>6. Then top up to
mark of the volumetric flask carefully.
</span></span>
7. As the final step prepared solution should be labelled.
Answer:
68133080.02 g
Explanation:
I believe that the question is to find the mass of air in the room and not the molar mass of air since the molar mass of air was already given in the question as 28.97 g/mol.
Now, if 1 mole of a gas occupies 22.4 L
x moles of air occupies 52,681,428.8 Liters
x = 1 * 52,681,428.8 /22.4
x = 2351849.5 moles of air
Now, number of moles = mass/ molar mass
but molar mass = 28.97 g/mol
2351849.5 = mass/28.97
mass = 2351849.5 * 28.97
mass = 68133080.02 g
