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Ede4ka [16]
3 years ago
9

Need help please omg Perform each of the following conversions being sure to set up the appropriate conversion factor in each ca

se 59. a. 12.5 in to centimeters c. 2513ft to miles 60. a. 2.23m to yards. c. 292cm to inches Please show work
Chemistry
1 answer:
sertanlavr [38]3 years ago
6 0

Answer:

The answer to your question is:

a) 31.75 cm

b) 0.475 miles

c) 2.44 yards

d) 11496.04 inches

Explanation:

Convert

a)           12.5 in to cm

         1 in ------------------- 2.54 cm

       12.5 in ----------------    x

            x = 12.5(2.54)/1 = 31.75/ = 31.75 cm

b)          2513 ft to miles

           1 mile -------------- 5280 ft

           x miles ------------ 2513 ft

        x = 2513(1)/5280 = 0.475 miles

c) 2.23 m to yards

              1 m -------------   1,094 yards

            2.23 m ----------     x

       x= 2.23x1.094/1 = 2.44 yards

d)  292 m to inches

            1 m ---------------- 39.37 inches

          292 m -------------     x

    x = 292 x 39.37/1 = 11496.04 inches

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First we have to calculate the moles of NaN_3

\text{Moles of }NaN_3=\frac{\text{Mass of }NaN_3}{\text{Molar mass of }NaN_3}

Molar mass of NaN_3 = 65.01 g/mole

\text{Moles of }NaN_3=\frac{20.2g}{65.01g/mole}=0.311mole

Now we have to calculate the moles of nitrogen gas.

The balanced chemical reaction is,

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

From the balanced reaction we conclude that

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So, 0.311 moles of NaN_3 react to give \frac{0.311}{2}\times 3=0.466 moles of N_2

Now we have to calculate the volume of nitrogen gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of N_2 gas = 1.00 atm

V = Volume of N_2 gas = ?

n = number of moles N_2 = 0.466 mole

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

T = Temperature of N_2 gas = 22^oC=273+22=295K

Putting values in above equation, we get:

1.00atm\times V=0.466mole\times (0.0821L.atm/mol.K)\times 295K

V=11.3L

As initially no nitrogen was present. So,

Volume expanded = Volume of nitrogen evolved

Thus,

Expansion work = Pressure × Volume

Expansion work = 1.00 atm × 11.3 L

Expansion work = 11.3 L.atm

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

Expansion work = 11.3 × 101.3 = 1144.69 J

Therefore, the value of work done for the system is 1144.69 J

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