So the ideal gas law is pv=nrt
The letter n stands for the number of moles. divide both side by rt to isolate.
pv/rt=n
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Answer
is: volume is 20 mL.<span>
c</span>₁(CH₃COOH) = 2,5 M.<span>
c</span>₂(CH₃COOH) = 0,5 M.<span>
V</span>₂(CH₃COOH) = 100 mL.<span>
V</span>₁(CH₃COOH) = ?<span>
c</span>₁(CH₃COOH) · V₁(CH₃COOH)
= c₂(CH₃COOH) · V₂(CH₃COOH).<span>
2,5 M · V</span>₁(CH₃COOH)
= 0,5 M · 100 mL.<span>
V</span>₁(CH₃COOH) = 0,5 M · 100 mL ÷ 2,5 M.<span>
V</span>₁(CH₃COOH) = 20 mL ÷ 1000 mL/L =0,02 L.
Answer:
As you move across a period, the atomic mass increases because the atomic number also increases. ... The atomic mass for any given atom mainly comes from the mass of the protons and neutrons.
Explanation:
Lets assume x volume of NaOH and x volume of HCl are added together.
NaOH ---> Na⁺ + OH⁻
NaOH is a strong base therefore it completely ionizes and releases OH⁻ ions into the medium
HCl ---> H⁺ + Cl⁻
HCl is a strong base and completely ionizes and releases H⁺ ions in to the medium. number of NaOH moles in 1 L - 0.1 mol
Therefore in x L - 0.1 /1 * x = 0.1x moles of NaOH present
Similarly in HCl x L contains - 0.1x moles of HCl
H⁺ + OH⁻ ---> H₂O
Due to complete ionisation, 0.1x moles of H⁺ ions and 0.1x moles of OH⁻ ions react to form 0.1x moles of H₂O. Therefore all H⁺ and OH⁻are completely used up and yield water molecules.
Then at this point the H⁺ and OH⁻ ions in the medium come from the weak dissociation of water. This is equivalent to 1 x 10⁻⁷M
pH = -log [H⁺]
pH = -log [10⁻⁷]
pH = 7
pH is therefore equals to 7 which means the solution is neutral
Answer:
5.41 g
Explanation:
Considering:
Or,
Given :
For tetraphenyl phosphonium chloride :
Molarity = 33.0 mM = 0.033 M (As, 1 mM = 0.001 M)
Volume = 0.45 L
Thus, moles of tetraphenyl phosphonium chloride :
Moles of TPPCl = 0.01485 moles
Molar mass of TPPCl = 342.39 g/mol
The formula for the calculation of moles is shown below:
Thus,
Mass of TPPCl = 5.0845 g
Also,
TPPCl is 94.0 % pure.
It means that 94.0 g is present in 100 g of powder
5.0845 g is present in 5.41 g of the powder.
<u>Answer - 5.41 g</u>
