Hello!
Ok so for this problem we use the ideal gas law of PV=nRT and I take it that the scientist needs to store 0.400 moles of gas and not miles.
So if we have
n=0.400mol
V=0.200L
T= 23degC= 273k+23c=296k
R=ideal gas constant= 0.0821 L*atm/mol*k
So now we rearrange equation for pressure(P)
P=nRT/V
P=((0.400mol)*(0.0821 L*atm/mol*k)*(296k))/(0.200L) = 48.6 atm of pressure
Hope this helps you understand the concept and how to solve yourself in the future!! Any questions, please feel free to ask!! Thank you kindly!!!
Answer:
The ratio of HC2H3O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) to HC2H3O2(aq) in the flask after the addition of 1.0mL of NaOH(aq) is 15 : 19 .
Explanation:
HC2H3O2 is CH₃⁻ COOH, which is also known as Acetic acid.
IUPAC name of this compound is Ethanoic acid.
Acetic acid has a basicity of 1. so there is one acidic hydrogen is acetic acid.
Given that, equivalence point was reached when 20.0mL of NaOH is added.
let the normality of acetic acid is N₁ and that of NaOH is N₂.
volume of acetic acid is V₁ and that of NaOH is V₂.
Equivalence point occurs when, N₁ × V₁ = N₂ × V₂.
⇒ N₁ × V₁ = N₂ × 20.
after the addition of 5.0mL of NaOH(aq), remaining N₁ × V° = N₂ × (20 - 5).
= N₂ × 15.
after the addition of 1.0mL of NaOH(aq), remaining N₁ × Vˣ = N₂ × (20 - 1).
= N₂ × 19.
⇒ V° : Vˣ = 15 : 19 .
⇒
Answer:
3.50*10^-11 mol3 dm-9
Explanation:
A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential difference between the rod and the SHE is 0.5812 V, the rod being positive. Calculate the solubility product constant for silver oxalate.
Ag2C2O4 --> 2Ag+ + C2O4 2-
So Ksp = [Ag+]^2 * [C2O42-]
In 1 L, 2.06*10^-4 mol of silver oxalate dissolve, giving, the same number of mol of oxalate ions, and twice the number of mol (4.12*10^-4) of silver ions.
So Ksp = (4.12*10^-4)^2 * (2.06*10^-4)
= 3.50*10^-11 mol3 dm-9
The answer is: [D]: " 417 cm³ " .
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Explanation: Use the formula:
V₁ /T₁= V₂ /T₂ ;
in which: V₁ = initial volume = 556 cm³ ;
T₁ = initial temperature = 278 K ;
V₂ = final ("new") temperature = 308 K
T₂ = final ("new:) volume = ?
Solve for "V₂" ;
Since: V₁ /T₁= V₂ /T₂ ;
We can rearrange this "equation/formula" to isolate "V₂" on one side of the equation; and then we can plug in our know values to solve for "V₂" ;
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V₁ /T₁= V₂ /T₂ ; Multiply EACH side of the equation by "T₂ " :
→ T₂ (V₁ /T₁) = T₂ (V₂ /T₂) ;
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to get:
↔ T₂ (V₂ /T₂) = T₂ (V₁ /T₁) ;
→ V₂ = T₂ (V₁ /T₁) ;
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Now, plug in our known values, to solve for "V₂" ;
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→ V₂ = T₂ (V₁ /T₁) ;
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→ V₂ = 308 K ( 556 cm³ /278 K) ;
→ The units of "K" cancel to "1" ; and we have:
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→ V₂ = 308*( 556 cm³ / 278 ) = [(208 * 556) / 278 ] cm³ ;
Note: We will keep the units of volume as: "cm³ "; since all the answer choices given are in units of: "cm³ " ; {that is, "cubic centimeters"}.
→ [(208 * 556) / 278 ] cm³ = [ (115,648) / (278) ] cm³ ;
→ For the "(115,648)" ; round to "3 (three significant figures)" ;
→ "(115,648)" → rounds to: "116,000" ;
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→ (116,000) / (278) = 417.2661870503597122 ;
→ round to 3 significant figures; → "417 cm³ " ;
→ which corresponds with "choice [D]".
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The answer is: [D]: "417 cm³ " .
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