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jekas [21]
3 years ago
10

A steel calorimeter has a volume of 75.0 mL and is charged with Oxygen gas to a

Chemistry
1 answer:
EastWind [94]3 years ago
7 0

Answer:

7.78×10¯³ mole

Explanation:

From the question given above, the following data were obtained:

Volume (V) = 75 mL

Pressure (P) = 255 kPa

Temperature (T) = 22.5 °C

Number of mole (n) =?

Next, we shall convert 75 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

75 mL = 75 mL × 1 L / 1000 mL

75 mL = 0.075 L

Next, we shall convert 22.5 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Temperature (T) = 22.5 °C

Temperature (T) = 22.5 °C + 273

Temperature (T) = 295.5 K

Finally, we shall determine the amount of oxygen molecules present in steel calorimeter. This can be obtained as follow:

Volume (V) = 0.075 L

Pressure (P) = 255 kPa

Temperature (T) = 295.5 K

Gas constant (R) = 8.314 KPa.L/Kmol

Number of mole (n) =?

PV = nRT

255 × 0.075 = n × 8.314 × 295.5

19.125 = n × 2456.787

Divide both side by 2456.787

n = 19.125 / 2456.787

n = 7.78×10¯³ mole

Thus, amount of oxygen molecules present in steel calorimeter is 7.78×10¯³ mole

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Answer:

pH = 13.1

Explanation:

Hello there!

In this case, according to the given information, we can set up the following equation:

HCl+KOH\rightarrow KCl+H_2O

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol

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3 0
3 years ago
Be sure to answer all parts. Consider the reaction N2(g) + 3H2(g) → 2NH3(g)ΔH o rxn = −92.6 kJ/mol If 4.0 moles of N2 react with
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Answer : The work done is, 1.98\times 10^4J

Explanation :

The given balanced chemical reaction is:

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Moles on reactant side = Moles of N_2 + Moles of H_2

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Now we have to calculate the change in volume of gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of gas = 1.0 atm

V = Volume of gas = ?

n = number of moles of gas = 8 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas = 25^oC=273+25=298K

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w = work done

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Now put all the given values in the above formula, we get:

w=-p\Delta V

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conversion used : (1 L.atm = 101.3 J)

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