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3 years ago

How do you find the pressure in a container if you don't have the temperature?

1 answer:

7 0

It depends on the process.
Like for example if the process is isothermal(temperature is constant), you can use,
PV = constant or P1V1 = P2V2 where P1V1 are initial conditions and P2V2 are final.

For adiabatic process,

PV^gamma = constant or P1V1 ^gamma = P2V2 ^gamma.

where gamma = Cp
------
Cv
Cp = specific heat at constant pressure and Cv = specific at constant volume.

Value of Gamma will be given in question.
Hope this helps!

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Calculate the density of CO2 at a pressure of 685.0 torr and 41.0°C . R=0.0821 (L*atm)/(mol *K)

Keith_Richards [23]

T=41+273=314 k

M=(12)+(16×2)=44g/mol

d=PM/RT

d=685×44/0.0821×314

d=1169.15 g/L

5 0

3 years ago

4.00 g of O2 gas are in a sealed, 2.00 L gas canister at 22.0 °C what is the pressure inside this container (in atm)?

Volgvan

Answer:

1.51448 atms

Explanation:

6 0

3 years ago

I NEED ANSWER PLEEEAASSEE!!!

emmasim [6.3K]

The answer is B. This is because Sodium has 1 valence electron and Fluorine has 7 valence electrons. All elements want 8 valence electrons so they may be stale, like the noble gases are. Hope this helps.

3 0

3 years ago

Read 2 more answers

Suppose you are a chemical engineer working with a client who produces toothpaste.The client wants to add a blue stripe with min

Nikitich [7]

Answer:

Explanation:

I would ask him why would he want to add the blue stripe?

5 0

3 years ago

What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according

ladessa [460]

Answer:

There is 27.6 mL of a 0.200 M HNO3 solution required

Explanation:

Step 1: The balanced equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

This means for 1 mole Na2CO3 consumed, there is consumed 2 mole of HNO3 and there is produced 2 moles of NaNO3, 1 mole of CO2 and 1 mole of H2O

Step 2: Calculating moles of Na2CO3

moles of Na2CO3 =volume of Na2CO3 * Molarity of Na2CO3

moles of Na2CO3 = 27.6 *10^-3 * 0.1 M = 0.00276 moles

Step 3: Calculating moles of HNO3

In the balanced equation, we can see that for 1 mole of Na2CO3 consumed, there are consumed 2 moles of HNO3.

So for 0.00276 moles consumed of Na2CO3, there are consumed 0.00552 moles of HNO3.

This means 0.00276 moles of the base Na2CO3 would react with 0.00552 moles of the acid HNO3

Step 4: Calculating the volume of HNO3

volume of HNO3 = moles of HNO3 / Molarity of HNO3

volume of HNO3 = 0.00552 moles / 0.200 M = 0.0276 L

0.0276 L = 27.6 ml

There is 27.6 mL of a 0.200 M HNO3 solution required

4 0

3 years ago