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rosijanka [135]
3 years ago
6

To what temperature must a balloon, initially at 25°c and 2.00 l, be heated in order to have a volume of 6.00 l?

Chemistry
2 answers:
Kaylis [27]3 years ago
6 0
V1 = 2.00 L 
<span>T1 = 25 + 273 = 298 K </span>
<span>V2 = 6.00 L </span>
<span>T2 = ? </span>
<span>Assuming the pressure is to remain constant, then </span>
<span>V1/T1 = V2/T2 </span>
<span>T2 = T1V2/V1 = (298)(6)/(2) = 894 deg K</span>
svp [43]3 years ago
3 0

Answer: 894 K

Solution :

Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

V\propto T     (At constant pressure and number of moles)

The combined gas equation is,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1 = initial volume of gas = 2.00 L

V_2 = final volume of gas = 6.00 L

T_1 = initial temperature of gas = 25^oC=273+25=298K

T_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get the final temperature of the gas

\frac{2.00L}{298K}=\frac{6.00L}{T_2K}

T_2=894K

Therefore, the temperature must be 894 K in order for balloon to have a volume of 6.00 L.

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A tank of oxygen has a volume of 1650 L. The temperature of the gas inside is 35?C. If there are 9750 moles of oxygen in the tan
Paul [167]

Answer:

2192.64 PSI.

Explanation:

  • From the general law of ideal gases:

<em>PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the container in L (V = 1650 L).

n is the no. of moles of the gas in mol (n = 9750 mol).

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature of the gas in (T = 35°C + 273 = 308 K).

∴ P = nRT/V = (9750 mol)(0.082 L.atm/mol.K)(308 K)/(1650 L) = 149.2 atm.

  • <u><em>To convert from atm to PSI:</em></u>

1 atm = 14.696 PSI.

<em>∴ P = 149.2 atm x (14.696 PSI/1.0 atm) = 2192.64 PSI.</em>

4 0
3 years ago
During the combustion of methane (CH4), one mole of methane gas combines with two moles of oxygen gas (O2) to produce one mole o
astraxan [27]

Answer:

CH4(g) +2O2____ CO2(g) +2H2O (l)

3 0
3 years ago
Place these elements in order of LOWEST electronegativity to HIGHEST.
mrs_skeptik [129]

Answer:

Barium

Molebdenum

Cobalt

Nitrogen

Explanation:

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8 0
3 years ago
Read 2 more answers
5.36 liters of nitrogen gas are at STP. What would be the new volume if we increased the moles from 3.5 moles to 6.0 moles?
Aleksandr [31]

Answer:

V_2=9.20L

Explanation:

Hello there!

In this case, according to the given STP (standard pressure and temperature), it is possible for us to realize that the equation to use here is the Avogadro's law as a directly proportional relationship between moles and volume:

\frac{V_2}{n_2}= \frac{V_1}{n_1}

In such a way, given the initial volume and both initial and final moles, we can easily compute the final volume as shown below:

V_2= \frac{V_1n_2}{n_1} \\\\V_2=\frac{5.36L*6.0mol}{3.5mol}\\\\V_2=9.20L

Best regards!

3 0
3 years ago
Suppose that 0.410 mol of methane, CH4(g), is reacted with 0.560 mol of fluorine, F2(g), forming CF4(g) and HF(g) as sole produc
Ber [7]

Answer:

The balanced equation for this reaction will be

                            CH4 + 4F2    →  CF4 + 4HF

We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2

So using the unitary method we will get that

  • 1 mole of CH4 → 4 mole of 4 mole of fluorine
  • 0.41 mole of methane  →  4*0.41 = 1.64 mole of fluorine for complete reaction

but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.

  • 4 moles of  fluorine →  1 mole of CF4
  • 0.56 mole →  \frac{1}{4} * 0.56 = 0.14mole of CF4
  • 4 moles of fluorine →  4 moles of HF
  • 0.56 mole of fluorine →  0.56 mole of HF

now to find the heat released we have the formula as

DELTA H = n * Delta H of product - n *delta H of reactant

where n is the moles of the reactant and product.

note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.

8 0
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