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Komok [63]
2 years ago
12

A thermometer reads a pressure of 120 kPa at 0 ˚C. What is the temperature when the thermometer reads a pressure of 80 kPa? (Cel

sius + 273 = Kelvin) Use Gay Lussac's Formula to solve. 182 K 192 K 202 K 212 K
Chemistry
1 answer:
Mariulka [41]2 years ago
4 0

Answer:

T₂ = 182 K

Explanation:

Given that,

Initial pressure, P₁ = 120 kPa  

Initial temperature, T₁ = 0˚C = 273 K

We need to find the final temperature when the pressure is 80 kPa.

We know that, Gay Lussac's Formula is :

\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\\\\T_2=\dfrac{P_2T_1}{P_1}\\\\T_2=\dfrac{80\ kPa\times 273}{120\ kPa}\\\\T_2=182\ K

So, the new temperature is equal to 182 K.

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You have to prepare some 2 M solutions, with 10 g of solute in each. What volume of solution will you prepare, for each solute b
alexdok [17]

Answer:

             0.045 L  or 45 mL

Explanation:

Moles = Mass/M.Mass

Moles = 10 g / 109.94 g/mol

Moles = 0.09 moles

Also,

Molarity = Moles / Vol in L

Or,

Vol in L = Moles / Molarity

Vol in L = 0.09 mol / 2 mol/L

Vol in L = 0.045 L

8 0
2 years ago
Consider the following reaction 2 N2O(g) =&gt; 2 N2(g) + O2(g) rate = k[N2O]. For an initial concentration of N2O of 0.50 M, cal
den301095 [7]

Answer:

After 2.0 minutes the concentration of N2O is 0.3325 M

Explanation:

Step 1: Data given

rate = k[N2O]

initial concentration of N2O of 0.50 M

k = 3.4 * 10^-3/s

Step 2: The balanced equation

2N2O(g) → 2 N2(g) + O2(g)  

Step 3: Calculate the concentration of N2O after 2.0 minutes

We use the rate law to derive a time dependent equation.

-d[N2O]/dt = k[N2O]

ln[N2O] = -kt + ln[N2O]i

 ⇒ with k = 3.4 *10^-3 /s

⇒ with t = 2.0 minutes = 120s

⇒ with [N2O]i = initial conc of N2O = 0.50 M

ln[N2O] = -(3.4*10^-3/s)*(120s) + ln(0.5)

ln[N2O] = -1.101

e^(ln[N2O]) = e^(-1.1011)

[N2O} = 0.3325 M

After 2.0 minutes the concentration of N2O is 0.3325 M

3 0
3 years ago
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