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3 years ago

A tank of gas is found to exert 8.6 atm at 38°C. What would be the required

Chemistry

1 answer:

Vesna [10]3 years ago

6 0

Answer:

36.2 K

Explanation:

Step 1: Given data

Initial pressure of the gas (P₁): 8.6 atm

Initial temperature of the gas (T₁): 38°C

Final pressure of the gas (P₂): 1.0 atm (standard pressure)

Final temperature of the gas (T₂): ?

Step 2: Convert T₁ to Kelvin

We will use the following expression.

K = °C +273.15

K = 38 °C +273.15 = 311 K

Step 3: Calculate T₂

We will use Gay Lussac's law.

P₁/T₁ = P₂/T₂

T₂ = P₂ × T₁/P₁

T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K

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svlad2 [7]

Answer:

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Explanation:

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3 years ago

52.1 mL of aqueous 0.255 M Pb(NO3)2 is mixed with 38.5 mL of 0.415 M NaCl. The equation for the precipitate reaction is: Pb(NO3)

Elan Coil [88]

Answer:

0.0585 M

Explanation:

Pb(NO₃)₂ (aq) + 2NaCl (aq) → PbCl₂ (s) + 2NaNO₃ (aq)

First we calculate the inital number of moles of each reagent, using the given volumes and concentrations:

0.255 M Pb(NO₃)₂ * 52.1 mL = 13.3 mmol Pb(NO₃)₂

0.415 M NaCl * 38.5 mL = 16.0 mmol NaCl

Then we calculate how many Pb(NO₃)₂ moles reacted with 16.0 mmoles of NaCl, using the stoichiometric coefficients of the reaction:

16.0 mmol NaCl * $\frac{1mmolPb(NO_3)_2}{2mmolNaCl}$ = 8.00 mmol Pb(NO₃)₂

Now we calculate the remaining number of Pb(NO₃)₂ moles after the reaction:

13.3 mmol - 8.00 mmol = 5.30 mmol Pb(NO₃)₂

Finally we divide the number of moles by the final volume to calculate the concentration:

5.30 mmol / (52.1 mL + 38.5 mL) = 0.0585 M

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3 years ago

Give the complete ionic equation for the reaction (if any) that occurs when aqueous solutions of lithium sulfide and copper (II)

Aloiza [94]

Answer:

$2Li^+_{(aq)}+S^{2-}_{(aq)}+Cu^{2+}_{(aq)}+2NO_3^{-}_{(aq)}\rightarrow CuS_{(s)}+2Li^+_{(aq)}+2NO_3^{-}_{(aq)}$

Explanation:

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

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$Li_2S_{(aq)}+Cu(NO_3)_2_{(aq)}\rightarrow CuS_{(s)}+2LiNO_3_{(aq)}$

The complete ionic equation in separated aqueous solution will be,

$2Li^+_{(aq)}+S^{2-}_{(aq)}+Cu^{2+}_{(aq)}+2NO_3^{-}_{(aq)}\rightarrow CuS_{(s)}+2Li^+_{(aq)}+2NO_3^{-}_{(aq)}$

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3 years ago