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azamat
3 years ago
5

a closed flask of air (0.250 L) contains 5.00 "puffs" of particles. The pressure probe on the flask reads 93 kPa. A student uses

a syringe to add an additional 3.00 "puffs" of air through the stopper. Find a new pressure inside the flask.
Chemistry
1 answer:
Sergio039 [100]3 years ago
3 0

Answer: New pressure inside the flask would be 148.8 kPa.

Explanation: The combined gas law equation is given by:

PV=nRT

As the flask is a closed flask, so the volume remains constant. Temperature is constant also.

So, the relation between pressure and number of moles becomes

P=n\\or\\\frac{P}{n}=constant

\frac{P_1}{n_1}=\frac{P_2}{n_2}

  • Initial conditions:

P_1=93kPa\\n_1=5\text{ puffs}

  • Final conditions: When additional 3 puffs of air is added

P_2=?kPa\\n_2=8\text{ puffs}

Putting the values, in above equation, we get

\frac{93}{5}=\frac{P_2}{8}\\P_2=148.8kPa

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Le Chatelier's principle states that when a change is brought to a system in equilibrium, the equilibrium will shift in a manner to reverse that change.

If the pressure is increased, the system will try to reduce the pressure. The only way it can do this is by producing less gas. Therefore, shifting the equilibrium to the left. Thus, the statement is true.
8 0
3 years ago
Find the linear function Cequals​f(F) that gives the reading on the Celsius temperature scale corresponding to a reading on the
padilas [110]

Answer:

C = (5/9) F - (160/9)

They both read equal at Z = - 40

Explanation:

We are looking for a linear function so we can write the following condition

Y = aX + b

Applying it to the exercise we got C = a F + b

Let's use the facts that C = 0 when F = 32 and C = 100 when F = 212

0 = 32 a + b            (1)

100 = 212 a + b       (2)

From (1) b = - 32 a , when we replace this in (2) we obtain a = (5/9)

and b = - (5/9)32 = - 160/9

Finally the linear function is C = (5/9) F - (160/9)

Both readings are equal at a Z number so

Z = (5/9) Z - 160/9

(4/9) Z = -160/9 and Z = - 40

6 0
3 years ago
Starting with 1.5052g of BaCl2•2H2O and excessH2SO4, how many grams of BaSO4 can be formed?
muminat
1.5052g BaCl2.2H2O => 1.5052g / 274.25 g/mol = 0.0054884 mol
=> 0.0054884 mol Ba 
<span>This means that at most 0.0054884 mol BaSO4 can form since Ba is the limiting reagent. </span>
<span>0.0054884 mol BaSO4 => 0.0054884 mol * 233.39 g/mol = 1.2809 g BaSO4</span>
6 0
3 years ago
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3 years ago
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What is the concentration of kl solution if 20.68g of solute was added to enough water to form 100ml solution?
rewona [7]

Answer:

1.25 M

Explanation:

Step 1: Given data

Mass of KI (solute): 20.68 g

Volume of the solution: 100 mL (0.100 L)

Step 2: Calculate the moles of solute

The molar mass of KI is 166.00 g/mol.

20.68 g × 1 mol/166.00 g = 0.1246 mol

Step 3: Calculate the molar concentration of KI

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.1246 mol/0.100 L= 1.25 M

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3 years ago
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