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denis-greek [22]
3 years ago
14

How many milliliters of nitrogen, N2, would have to be collected at 99.19 kPa and 28oC to have a sample containing 0.015 moles o

f N2
Chemistry
1 answer:
Semmy [17]3 years ago
4 0

Answer:

378mL

Explanation:

The following data were obtained from the question:

Pressure (P) = 99.19 kPa

Temperature (T) = 28°C

Number of mole (n) = 0.015 mole

Volume (V) =...?

Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:

For Pressure:

101.325 KPa = 1 atm

Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm

For Temperature:

T(K) = T(°C) + 273

T(°C) = 28°C

T(K) = 28°C + 273 = 301K.

Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:

PV = nRT

Pressure (P) = 0.98 atm

Temperature (T) = 301K

Number of mole (n) = 0.015 mole

Gas constant (R) = 0.0821atm.L/Kmol.

Volume (V) =...?

0.98 x V = 0.015 x 0.0821 x 301

Divide both side by 0.98

V = (0.015 x 0.0821 x 301) /0.98

V = 0.378 L

Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:

1L = 1000mL

Therefore, 0.378L = 0.378 x 1000 = 378mL

Therefore, the volume of N2 collected is 378mL

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Fission is similar to mitosis because they both involve splitting. 

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Comment Both propane and benzene are hydrocarbons. As a rule,
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The enthalpy change : -196.2 kJ/mol

<h3>Further explanation  </h3>

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

Reaction

2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)

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5 0
3 years ago
1. An electric power plant uses energy from burning coal to generate steam at 450 °C. The plant is cooled by 20 °C water
kap26 [50]

Answer:

Explanation:

Efficiency of the electric power plant is e=1-\frac{T_{2}}{T_{1}}

Here Temperature of hot source T_{1} = 450^{o}C=450+273=723 K

and Temperature of sink T_{1} = 20^{o}C=20+273=293 K

Hence the efficiency is  e=1-\frac{293}{723}=0.5947=59.47%

Now another formula for thermal efficiency Is

e_{therm}=\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{W}{Q_{1}}

Here QI is the of heat taken from source 100 MJ ; Q2 of heat transferred to the sink (river) to be found

W is the of work done and W = QI -Q2

Hence Frome_{therm}=\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{W}{Q_{1}}

W=e(Q_{1})=(0.5947)(100)=59.47MJ

Hence the of heat transferred to the river Is Q_{2} -W = (100 -59.47=40.53

6 0
3 years ago
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