Answer:
The initial volume in mL is 5959.2 mL
Explanation:
As the number of moles of a gas increases, the volume also increases. Hence, number of moles and volumes are directly proportional i.e
n ∝ V
Where n is the number of moles and V is the volume
Then, n = cV
c is the proportionality constant
∴n/V = c
Hence n₁/V₁ = n₂/V₂
Where n₁ is the initial number of moles
V₁ is the initial volume
n₂ is the final number of moles
and V₂ is the final volume.
From the question,
n₁ = 0.693 moles
V₁ = ?
n₂ = 0.928 moles
V₂ = 7.98 L
Putting the values into the equation
n₁/V₁ = n₂/V₂
0.693 / V₁ = 0.928 / 7.98
Cross multiply
∴ 0.928V₁ = 0.693 × 7.98
0.928V₁ = 5.53014
V₁ = 5.53014/0.928
V₁ = 5.9592 L
To convert to mL, multiply by 1000
∴ V₁ = 5.9592 × 1000 mL
V₁ = 5959.2 mL
Hence, the initial volume in mL is 5959.2 mL
Answer:
b) 5.87 E23 molecules
Explanation:
∴ mm SO3 = 80.066 g/mol
⇒ molecules SO3 = (78.0 g)(mol/80.066 g)(6.022 E23 molec/mol)
⇒ molec SO3 = 5.866 E23 molecules SO3
Answer:
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Answer: im pretty sure its D
Explanation:
The concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.
What is meant by concentration?
Concentration is the total amount of solute present in the given volume of solution. this is expressed in terms of molarity, molality, mole fraction, normality etc. The term concentration mostly refers to the solvents and solutes present in the solution.
Concentration of hydroxide ions can be calculated by,
M (OH^-) = V (NaOH) x M (NaOH) / V (total) = 50ml x 0.250M / 50ml + 120ml = 0.0735M = 7.35 x 10^-2 M.
where M (OH^-) = concentration of hydroxide ions, V(NaOH) = volume of NaOH, M(NaOH) = concentration of NaOH.
Therefore, the concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.
To learn more about concentration click on the given link brainly.com/question/17206790
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